algebra
polynomial
polynomial division
Laurentiu Panaitopol
IMO
IMO 2002
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Ljunggren's theorem kills it.

//my bad, there is answer m=5, n=3.

May I know that what is"Ljunggren's theorem"?
Thanks.

[hide]

Suppose that we can write $ a^m\plus{}a\minus{}1\equal{}(a^n\plus{}a^2\minus{}1)Q(x)\plus{}R(x)$, with $ R(x)$ not identically zero. Then by the conditions, $ R(x)$ must have infinitely many roots, contradiction; hence, $ a^n\plus{}a^2\minus{}1|a^m\plus{}a\minus{}1$. Since $ 0^n\plus{}0^2\minus{}1<0$ and $ 1^n\plus{}1^2\minus{}1>0$, $ a^n\plus{}a^2\minus{}1\equal{}0$ has a root $ 0<r<1$. Then also $ r^m\plus{}r\minus{}1\equal{}0$, so $ r^m\plus{}r\equal{}r^n\plus{}r^2 \Rightarrow r(1\minus{}r)\equal{}r^n(1\minus{}r^{m\minus{}n}) \Rightarrow r(1\minus{}r)\equal{}(1\minus{}r^2)(1\minus{}r^{m\minus{}n})$. Continuing, we have that $ r\equal{}1\minus{}r^{m\minus{}n}\plus{}r\minus{}r^{m\minus{}n\plus{}1} \Rightarrow r^{m\minus{}n\plus{}1}\plus{}r^{m\minus{}n}\equal{}1$. Since $ r<1$, we must have $ m\minus{}n\plus{}1\le n$ or else $ r^{m\minus{}n\plus{}1}\plus{}r^{m\minus{}n}< r^n\plus{}r^{n\minus{}1}\le r^n\plus{}r^2\equal{}1$, contradiction. Hence, $ m\le 2n\minus{}1$.

Clearly, $ a^n\plus{}a^2\minus{}1|(a^{m\minus{}n}(a^n\plus{}a^2\minus{}1)\minus{}(a^m\plus{}a\minus{}1)) \Rightarrow a^n\plus{}a^2\minus{}1|a^{m\minus{}n\plus{}2}\minus{}a^{m\minus{}n}\minus{}a\plus{}1$. Since $ a^{m\minus{}n\plus{}2}\minus{}a^{m\minus{}n}\minus{}a\plus{}1$ is not identically zero, and since $ m\le 2n\minus{}1$, $ m\minus{}n\plus{}2$ is either $ m$ or $ m\plus{}1$. In the first case, we must have $ a^{m\minus{}n\plus{}2}\minus{}a^{m\minus{}n}\minus{}a\plus{}1\equal{}a^n\plus{}a^2\minus{}1$, but this is impossible. In the second case, we must have $ (a\minus{}1)(a^n\plus{}a^2\minus{}1)\equal{} a^{n\plus{}1}\minus{}a^{n\minus{}1}\minus{}a\plus{}1 \Rightarrow \minus{}a^{n\minus{}1}\equal{}\minus{}a^n\plus{}a^3\minus{}a^2$, which can only happen if $ n\equal{}3$, and therefore $ m\equal{}5$. As $ (a^3\plus{}a^2\minus{}1)(a^2\minus{}a\plus{}1)\equal{}(a^5\plus{}a\minus{}1)$, this solution works. So the only answer is $ (m,n)\equal{}(5,3)$.

[/hide]

[quote="hmm"][hide]

Suppose that we can write $ a^m \plus{} a \minus{} 1 \equal{} (a^n \plus{} a^2 \minus{} 1)Q(x) \plus{} R(x)$, with $ R(x)$ not identically zero. Then by the conditions, $ R(x)$ must have infinitely many roots, contradiction; hence, $ a^n \plus{} a^2 \minus{} 1|a^m \plus{} a \minus{} 1$. Since $ 0^n \plus{} 0^2 \minus{} 1 < 0$ and $ 1^n \plus{} 1^2 \minus{} 1 > 0$, $ a^n \plus{} a^2 \minus{} 1 \equal{} 0$ has a root $ 0 < r < 1$. Then also $ r^m \plus{} r \minus{} 1 \equal{} 0$, so $ r^m \plus{} r \equal{} r^n \plus{} r^2 \Rightarrow r(1 \minus{} r) \equal{} r^n(1 \minus{} r^{m \minus{} n}) \Rightarrow r(1 \minus{} r) \equal{} (1 \minus{} r^2)(1 \minus{} r^{m \minus{} n})$. Continuing, we have that $ r \equal{} 1 \minus{} r^{m \minus{} n} \plus{} r \minus{} r^{m \minus{} n \plus{} 1} \Rightarrow r^{m \minus{} n \plus{} 1} \plus{} r^{m \minus{} n} \equal{} 1$. Since $ r < 1$, we must have $ m \minus{} n \plus{} 1\le n$ or else $ r^{m \minus{} n \plus{} 1} \plus{} r^{m \minus{} n} < r^n \plus{} r^{n \minus{} 1}\le r^n \plus{} r^2 \equal{} 1$, contradiction. Hence, $ m\le 2n \minus{} 1$.
[/hide][/quote]
Wow, that was quite ingenious. :D I have another finish to this problem which I feel is slightly cleaner:

Since $ a^n \plus{} a^2 \minus{} 1 | a^m \plus{} a \minus{} 1$, $ a^n \plus{} a^2 \minus{} 1 | (a\plus{}1)(a^m \plus{} a \minus{} 1) \minus{} (a^n \plus{} a^2 \minus{} 1) \equal{} (a^{m\plus{}1} \plus{} a^m) \plus{} (a^2 \minus{} 1) \minus{} (a^n \plus{} a^2 \minus{} 1) \equal{} a^{m\plus{}1} \plus{} a^m \minus{} a^n \equal{} a^n(a^{m\minus{}n\plus{}1} \plus{} a^{m\minus{}n} \minus{} 1)$. Therefore, $ a^{n} \plus{} a^2 \minus{} 1 | a^{m\minus{}n\plus{}1} \plus{} a^{m\minus{}n} \minus{} 1$. Hence, $ n \leq m \minus{} n \plus{} 1$, so $ 2n \minus{} 1 \leq m$. But it was  have shown that $ m \leq 2n \minus{} 1$, so we have that $ m \equal{} 2n \minus{} 1$. Therefore, $ a^n \plus{} a^2 \minus{} 1 | a^n \plus{} a^{n\minus{}1} \minus{} 1$, so $ a^n \plus{} a^2 \minus{} 1 | a^{n\minus{}1} \minus{} a^2$. The only way this can be is if $ a^{n\minus{}1} \equal{} a^2$, yielding $ n \equal{} 3$ and $ m \equal{} 5$, as desired.

We first show that $f(a) = a^n+a^2-1$ has no double root. But it is easy to see that $\gcd(f(a), f'(a)) = \gcd(a^n+a^2-1, na^{n-1}+2a) = 1$ (not hard, I can show this more explicitly). Now let $r$ be a root of $f(a)$. Then $r^m+r-1 = 0, r^n+r^2-1 = 0$.

Rearranging the above equations gives $r^m = 1-r, r^n = (1-r)(1+r)$. Dividing these equations and rearranging gives $r^{m-n+1}+r^{m-n}-1 = 0$. But this holds for all roots $r$, so we get that $f(a) \vert a^{k+1}+a^k-1$, where $k = m-n$. By a rather famous result, $X^n-X-1$ is irreducible for all $n$, so if we reverse the coefficients, $X^n+X^{n-1}-1$ is also irreducible. Therefore, $f(a) = a^{k+1} + a^k - 1 \implies n = 3, k = 2, \implies m = 5, n = 3$.

This [url=http://www.mscand.dk/article.php?id=1472]article[/url] seems to contain the proof that $X^n-X-1$ is irreducible in the beginning and much more later.

My solution is written in the language of modules of polynomials, though the key ideas are nearly the same. It is less elegant than solutions above (As usual). It may appear lengthy but it is just direct computations ( I didn't omit any details, it is just the same as in my scratch paper)

Denote $f(a)=\frac{a^m+a-1}{a^n+a^2-1}$. . We let $P_m(x)=x^m+x-1$ and $Q_n(x)=x^n+x^2-1$, we have since both polynomials are monics, we can use [i]Euclidean theorem for monic integer polynomials[/i], to conclude that there is $S,R\in \mathbb{Z}[x]$ where $ \deg R < n$, such that $ P_m(x)= Q_m(x) S(x) + R(x)$, we have then \[ f(x) = \frac{x^m+x-1}{x^n+x^2-1} = \frac {P_m(x)}{Q_n(x)}= S(x) + \frac{R(x)}{Q_n(x)} \] If $R(x)$ is non-zero, We get the $\frac{R(x)}{Q_n(x)} \rightarrow 0$, which imply that for all but finitely many integers $s$, $ |\frac{R(x)}{Q_n(x)}| <1 $. And as that term never achieves zero for large numbers, we can't have $f(a)$ integer for infinitely many $a$'s. So we must have $Q_n | P_m$ in $\mathbb{Z}[x]$. Easily one can  see that we must have $m > n$ .

Now we prove that $Q_n | P_m$ iff $(m,n)=(5,3)$. We do module arithmetic in the Euclidean domain $\mathbb{Q}[x]$ to prove that. Note that $(x,Q_n(x))=1$ and $(x-1,Q_n(x))=1$ in $\mathbb{Q}[x]$, and hence \[ \begin{array}{lcl}
x^m+x-1 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^m+x-x^n-x^2 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^{m-1}+1-x^{n-1}-x & \equiv  & 0\pmod{x^n+x^2-1}  \\
\iff x^{n-1}(x^{m-n} - 1) + (1-x) & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^{m-2} + \cdots + x^{n-1} -1 & \equiv &0 \pmod{x^n+x^2-1}
\end{array}\]
If we let $m=n+i$ we get
\begin{align*}
x^{m-2} + \cdots + x^{n-1} -1 &\equiv x^{n+i-2}+ \cdots + x^{n-1} + x^{i} + x^{i-1} + \cdots + x^2 -1   \\
&\equiv  (x^{n+i-2}+ x^{i})+ (x^{n+i-3}+x^{i-1}) + \cdots + ( x^{n} + x^{2}) + x^{n-1} - x^{i} - x^{i-1} - \cdots - x^{2} -1 \\
&\equiv x^{i-2}+ x^{i-3}+ \cdots +1 + x^{n-1} - x^{i} - x^{i-1} - \cdots - x^{2} -1 \\
&\equiv x^{n-1} + x - x^{i} - x^{i-1} \pmod{x^n+x^2-1}
\end{align*}
Now if $i<n$ we get $x^{n-1} + x - x^{i} - x^{i-1} = 0 $ which cannot happen unless $n-1=i$ and $i-1=1$ and hence $(n,i)=(3,2)$ and so we have the solution $(m,n) = (5,3)$. Now if $i=n+j ; j\geq0$ , then \[ \begin{array}{lcl}
x^{n-1} + x - x^{i} - x^{i-1} & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff - x^{n-1} - x + x^{n+j} + x^{n+j-1} & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff  x^{n+j-1} + x^{n+j-2} - x^{n-2} - 1 & \equiv  & 0\pmod{x^n+x^2-1}  \\
\iff  x^{n-2}(x^{j+1}-1) + x^{n+j-2} - 1 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff     x^{n+j-2} + 2x^{n+j-3} + \cdots + 2x^{n-2}+ x^{n} + \cdots + 1 & \equiv &0 \pmod{x^n+x^2-1}
\end{array}\]

But $x^{n+j-2} + 2x^{n+j-3} + \cdots + 2x^{n-2}+ x^{n} + \cdots + 1 \geq 1$ for $ x\geq0$ , while $s^n+s^2-1=0$ for some $s\in(0,1)$ since divisibility applies also in $\mathbb{R}[x]$ , we get $s^{n+j-2} + 2s^{n+j-3} + \cdots + 2s^{n-2}+ s^{n} + \cdots + 1 = 0$, getting a contradiction.

I found a pretty cool and unique solution to this problem. Can someone check for its validity?

[hide=Solution]
Let $x^m+x-1=(x^n+x^2-1)Q(x)+R(x)$, where $Q,R\in\mathbb{Z}[x]$ and $\deg(R)<m$. Dividing both sides by $x^n+x^2-1$, we have that $\frac{x^m+x-1}{x^n+x^2-1}=Q(x)+\frac{R(x)}{x^n+x^2-1}$, so $\frac{R(a)}{a^n+a^2-1}$ must be an integer for infinitely many positive integers $a$. Since $\deg(R)<n$, for all sufficiently large values of $x$ we have that $\left|\frac{R(x)}{x^n+x^2-1}\right|<1$, so this forces $R(x)=0$. Hence, $x^n+x^2-1\mid x^n+x-1$ is a polynomial identity.

Now, we work in the unique factorization domain $\mathbb{Z}[\sqrt2]$. Since $x^n+x^2-1\mid x^n+x-1$ is a polynomial identity, we plug in $x=\sqrt2$ to get $(\sqrt2)^n+1\mid(\sqrt2)^m+\sqrt2-1$ in $\mathbb{Z}[\sqrt2]$.

If $n$ is even, then $(\sqrt2)^n+1$ is an integer, so it must divide $(\sqrt2)^m+\sqrt2-1$, which is of the form $(2^a+1)\sqrt2-1$ or $\sqrt2+(2^a-1)$. In either case, we have that $(\sqrt2)^n+1\mid1$, a contradiction since $(\sqrt2)^n+1>1$.

If $n$ is odd, let $m=2a+1$ and $k$ be the remainder when $n$ is divided by $m$. By reducing $(\sqrt2)^m+\sqrt2-1$ mod $(\sqrt2)^n+1$, we have that $2^a\sqrt2+1\mid (-1)^i(\sqrt2)^k+\sqrt2-1$ for some integer $i$. Now, we split into two cases:

If $k$ is even, then let $k=2b$. Note that $a\geq b$. We have that $a^2\sqrt2+1\mid\sqrt2+((-1)^i2^b-1)$. The absolute value of the norm of $2^a\sqrt2+1$ is $2\cdot(2^a)^2-1$ and the absolute value of the norm of $\sqrt2+((-1)^i2^b-1)$ is of the form $(2^b\pm1)^2-2$. If $a>1$, then $2^a>2$, so $2\cdot(2^a)^2-1>(2^a+1)^2-2\geq(2^b\pm1)^2-2$, so this forces $a=1$ and thus $n=3$.

If $k$ is odd, then let $k=2b+1$. Note that $a>b$. We have that $2^a\sqrt2+1\mid((-1)^i2^b+1)\sqrt2-1$. The absolute value of the norm of $2^a\sqrt2+1$ is $2\cdot(2^a)^2-1$ and the absolute value of the norm of $((-1)^i2^b+1)\sqrt2-1$ is of the form $2(2^b\pm1)^2-1$. But we clearly have $2^a\geq2^b+1$ with equality iff $a=1$ and $b=0$, so this forces $n=3$.

So now we want to solve $x^3+x^2-1\mid x^m+x-1$ for $m$. Suppose that this had two distinct solutions $m=m_1$ and $m=m_2$ with $m_1>m_2$. Then, $x^3+x^2-1\mid(x^{m_1}+x-1)-(x^{m_2}+x-1)=x^{m_2}(x^{m_1-m_2}-1)$. But this is impossible because $x^3+x^2-1$ is irreducible and is not $x$ or a cyclotomic polynomial. Hence, there is at most one solution. Note that $m=5$ works, with $x^5+x-1=(x^3+x^2-1)(x^2-x+1)$, so our only solution is $(5,3)$.
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Notice that $m>n$ since $m,n>3$ implies that if $m \leq n$, then $0<\frac{a^m+a-1}{a^n+a^2-1}<1$ for $a$ relatively large. First we claim that $\frac{a^m+a-1}{a^n+a^2-1}$ must always be an integer. Indeed, since $\frac{a^m+a-1}{a^n+a^2-1}=R(x)+\frac{Q(x)}{a^n+a^2-1}$ for polynomials $R,Q$, with $deg(Q)<deg(a^n+a^2+1)$, implying that for large $a$, $Q(x)<a^n+a^2-1$ occurs, meaning that the quotient cannot yield an integer unless $Q(x)=0$ for infinitely many values, implying the above claim. Now we can bound. Since
$a^n + a^2 - 1 | a^m + a - 1$,  we get that $ a^n + a^2 - 1 | (a+1)(a^m + a - 1) - (a^n + a^2 - 1)$.
But $(a+1)(a^m + a - 1) - (a^n + a^2 - 1)= a^{m+1} + a^m - a^n = a^n(a^{m-n+1} + a^{m-n} - 1)$.
Therefore, $ a^{n} + a^2 - 1 | a^{m-n+1} + a^{m-n} - 1$. Hence, $ n \leq m - n + 1$, so $ 2n - 1 \leq m$. Plugging in $a=2$, we immediately get $2^n+3|2^m+1$ with $m<2n$. Then $gcd(2^n+3,2^m+1)=gcd(3(2^{m-n})-1,2^n+3)$. Since $3(2^{m-n})-1<3(2^{m-n})<3(2^n)$ and $3(2^{m-n})-1$ is odd, $3(2^{m-n})-1=2^n+3$. This rearranges to $3(2^k)=2^n+4$ for $k=m-n$. Since $n>2$, $k<2$, implying the only solution to be $m=5,n=3$.

The condition is equivalent to $a^n+a^2-1$ dividing $a^m+a-1$ as polynomials. The big step is the following analytic one.

[b][color=red]Claim:[/color][/b]  	We must have $m \le 2n$.

[i]Proof.[/i]  	Assume on contrary $m > 2n$ 	and let $0 < r < 1$ be the unique real number 	with $r^n+r^2 = 1$, hence $r^m+r = 1$. 	But now 	\begin{align*} 		0 &= r^m + r - 1 < r(r^n)^2 + r - 1 = r\left( (1-r^2)^2+1 \right) - 1 \\ 		&= -(1-r)\left( r^4+r^3-r^2-r+1 \right). 	\end{align*} 	As $1-r > 0$ and $r^4+r^3-r^2-r+1 > 0$, this is a contradiction $\blacksquare$

Now for the algebraic part. Obviously $m > n$. \begin{align*} 	a^n+a^2-1 &\mid a^m+a-1 \\ 	\iff a^n+a^2-1 &\mid (a^m+a-1)(a+1) = a^m(a+1) + (a^2-1)  \\ 	\iff a^n+a^2-1 &\mid a^m(a+1) - a^n \\ 	\iff a^n+a^2-1 &\mid a^{m-n}(a+1) - 1. \end{align*} The right-hand side has degree $m-n+1 \le n+1$, and the leading coefficients are both $+1$. So the only possible situations are \begin{align*} 	a^{m-n}(a+1) - 1 &= (a+1)\left( a^n+a^2-1 \right) \\ 	a^{m-n}(a+1) + 1 &=  a^n+a^2-1. \end{align*} The former fails by just taking $a=-1$; the latter implies $(m,n) = (5,3)$. As our work was reversible, this also implies $(m,n) = (5,3)$ works, done.

[quote=orl]Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1}  \] is itself an integer.

[i]Laurentiu Panaitopol, Romania[/i][/quote]

It is standard to prove that $x^n+x^2-1 \mid x^m+x-1$ as polynomials, by the given condition. Now the former has different signs at $x=0$ and $x=1$, so pick a root $0<\alpha<1$. Notice that $\alpha^n+\alpha^2=1$ and $\alpha^m+\alpha=1$. Suppose $m \ge 2n$. Then $$\alpha^m \le \alpha^{2n} \implies 1=\alpha^2+\alpha^n<\alpha+\alpha^{2n} \iff \alpha^{2n}-\alpha^2>\alpha^n-\alpha \iff \alpha^n+\alpha<1=\alpha^2+\alpha^n \iff \alpha<\alpha^2 \iff \alpha>1$$ which is a contradiction!

So $m<2n$. Now $x^n+x^2-1 \mid x^{m-n+1}+x^{m-n}-1$ by taking residues. Finally, the latter polynomial has degree $m-n+1 \le (2n-1)-(n-1)$ so in fact, these two polynomials are equal, hence $m-n=2, m=2n-1$ so $m=5, n=3$. Clearly, this pair works, so we're done.

Kind of nice.
The condition forces the polynomial $a^n + a^2 - 1$ to divide $a^m + a - 1$.  \\
$\textbf{Claim 01.}$ $m < 2n$. \\
$\textit{Proof.}$ Suppose otherwise, we have $m \ge 2n$.
Notice that $(0)^n + (0)^2 - 1 = -1$ and $(1)^n + (1)^2 - 1 = 1$. By the Intermediate Value Theorem, we have that the polynomial $a^n + a^2 - 1$ has a root $\alpha$ which satisfy $0 < \alpha < 1$. Therefore,
\[ \alpha^m + \alpha = \alpha^n + \alpha^2 \Rightarrow \alpha (1 - \alpha) = \alpha^n (1 - \alpha^{m - n}) = (1 - \alpha^2)(1 - \alpha^{m - n}) \]
This gives us
\[ \alpha = (1 + \alpha)(1 - \alpha^{m - n}) \]
which can be rewritten as $\alpha^{m - n} + \alpha^{m - n + 1} = 1$. Since $m - n + 1 > n \ge 3$ and $0 < \alpha < 1$. We have
\[1 =  \alpha^{m - n} + \alpha^{m - n + 1} \le \alpha^n + \alpha^{n - 1} \le \alpha^n + \alpha^2= 1\]
results to a clear contradiction.

Now, we have
\[ x^n + x^2 - 1 | x^{m - n + 2} - x^{m - n} - x + 1 \]
Notice that we have the bound $n \le m - n + 2 \le n + 1$. \\
We'll prove that we must have $m - n + 2 = n + 1$.\\
Assume otherwise, then we must have $LHS = RHS$ since they are both the polynomial in the same degree and both having leading coefficient $1$, which is a clear contradiction as $RHS$ has constant $+1$ and $LHS$ has constant $-1$. \\
Then we have $m - n + 2 = n + 1$.
By considering the constant term ad the largest coefficient, we deduce
\[ (x^n + x^2 - 1)(x - 1) = x^{n + 1} - x^{n - 1} - x + 1 \]
But taking $x = 2$. We have $LHS = 2^n + 2^2 - 1 \le 2^{n} + 2^{n - 1} - 1$ since $n \ge 3$. \\
For equality to hold, we must have $n = 3$. This forces $m = 5$, and we could check that
\[ (x^3 + x^2 - 1)(x^2 - x + 1) = x^5 + x - 1 \]
Therefore, the pair $(3,5)$ clearly satisfies.

[b]Lemma:[/b] For $ a, m, n\in \mathbb{N} $, we have
\[a^n+a^2-1\ |\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right)\]

[b]Proof:[/b] We proceed by induction on $ i $. For $ i=0, 1 $, it is true. So assume for some $i>2$, it is true.

\begin{align*}
	a^n+a^2-1\ &|\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right)\\[1em]
	\implies a^n+a^2-1\ &|\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right) - (-1)^i(a^n+a^2-1)\\[.5em]
	&=a^{m-i} +  (-1)^i\left(-a^n+ a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^{i+1}(a^2-a)\\[1em]
	\therefore a^n+a^2-1\ &|\ a^{m-i-1} + (-1)^{i+1}\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^{i+1}\left(a-1\right)\\
\end{align*}
Which proves our lemma.

Now, let $ i=n $, we get,
\[a^n+a^2-1\ |\ a^{m-n}  + (-1)^n \left(\frac{a^n+1}{a+1}\right) + (-1)^n(a-1)\]
Clearly, $ m\ge n $. Let $ m=nk+q $. If $ n $ is even,
\begin{align*}
a^n+a^2-1\ &|\ a^{m-n}+a-1 + \left(\frac{a^n+1}{a+1}\right)\\[.5em]
\implies a^n+a^2-1\ &|\ a^q+a-1 + k\left(\frac{a^n+1}{a+1}\right)
\end{align*}

Which is not possible since the polynomial on the right side has degree at most $ n-1 $, and can't be $ 0 $ for all $ a\in \mathbb{Z} $.

So, $ n $ is odd. Then we have,

\begin{align*}
a^n+a^2-1\ &|\ a^{m-n} + a-1 - \left(\frac{a^n+1}{a+1}\right) - 2(a-1)\\[.5em]
\implies a^n+a^2-1\ &|\ a^q + a -1 - k\left(\frac{a^n+1}{a+1}\right) -2k(a-1) = P(a)
\end{align*}

Since $ P(a) $ has degree at most $ n-1 $, $ P(a) =0 $ for all $ a\in \mathbb{Z} $.

\begin{align*}
	P(a) &= a^q + a -1 - k\left(\frac{a^n+1}{a+1}\right) -2k(a-1)\\
	\therefore a^q + a - 1 &= k\left(\frac{a^n+1}{a+1}\right)+2k(a-1)
\end{align*}

Comparing the coefficients and degrees on both sides, we have, $ q=n-1=2,\ k=1 $. Which gives us the only solution $ (m, n) = (5, 3) $.

The problem is a joke if someone knows a particular lemma. ;)
Here's my [hide=solution]
We claim that only $(m,n)=(5,3)$ satisfy the problem conditions. Its easy to see that they work
Let $$f(x) \overset{\text {def}}{=} x^m+x-1$$ and
$$g(x) \overset{\text {def}}{=} x^n+x^2-1$$
Now pick a root $\alpha$ of $g(x)$ . [From the fact that $g(0) \cdot g(1) <0$, we know that $\alpha$ can be possibly real .].

We have $$\alpha ^m = 1-\alpha$$ and
$$\alpha ^n = (1-\alpha)(1+\alpha).$$

So we have $$\alpha^{m+1}+\alpha^{m} = \alpha ^n \implies \boxed{ \alpha^{m-n+1}+\alpha^{m-n}-1=0}$$

Since this holds $\forall \alpha$ such that $g(\alpha)=0$ , hence we may safely conclude that $$g(x) \mid x^{m-n+1}+x^{m-n}-1 \overset{\text {def}}{=} P(x)$$

Now by Selmer's lemma , we know that $\forall n\in \mathbb{N}$ ,
$$X^n-X-1 \text {   is irreducible in   } \mathbb{Z}[X]$$
$$\implies X^n+X^{n-1}-1 \text {   is irreducible in  } \mathbb{Z}[X]$$

Where the second polynomial is just the reciprocal polynomial of the first.

Hence , since $P(X)$ is irreducible, we conclude,

$$n=m-n+1$$
$$m-n=2$$ .

This gives us our desired solutions.  $\blacksquare$

[/hide]

Cool problem!

We require $a^n+a^2-1\mid a^m+a-1$, else the remainder (which is deg $n-1$ at most) will eventually be smaller than $a^n+a^2-1$ for large enough $a.$

Also note $n<2m$ otherwise $a^m+a>a^{2m}+a^2>a^n+a^2$ for $0<a<1$ and obviously one of the roots has this value. So then the remainder is $-a^{m-n+2}+a^{m-n}+a-1,$ so we want $a^n+a^2-1\mid a^{m-n+2}-a^{m-n}-a+1,$ which must be equal to $a^n+a^2-1$ or $(a^n+a^2-1)(a-1)$ as $m-n+2\leq m+1.$ For size reasons $m-n+2=m$ or $m-n+2=m+1$. The first case fails by comparison of $a$ coefficient, and the second case requires $n=3$, which gives us $(a^3+a^2-1)(a-1)=a^m+a-1,$ fixing $m=5.$

So just $(3,5)$ works.

[hide=Solution]
Suppose that $a^m+a-1=f(a)(a^n+a^2-1)+r(a)$ where the degree of $r$ is less than $n$. Then, there are infinitely many positive integers such that $\frac{r(a)}{a^n+a^2-1}$ is an integer, so $r(a)=0$. Therefore, $a^n+a^2-1\mid a^m+a-1$. Let $r$ be a root of $a^n+a^2-1$ between $0$ and $1$. Then, $r^m+r-1=0$. Since $a^n+a^2-1=a^2(a^{n-2}+1)-1$ and $a^m+a-1=a(a^{m-1}+1)-1$ are increasing, this means that $r$ is the only root of $a^n+a^2-1$ and $a^m+a-1$. Suppose that $m=y\ln(y)$. Then, we claim that $r>\frac{y-1}y$. We have
\begin{align*}
\left(\frac{y-1}y\right)^m-\frac1y&=\frac1{\left(1+\frac1{y-1}\right)^m}-\frac1y\\
&=\frac1{\left(1+\frac1{y-1}\right)^{y\ln(y)}}-\frac1y\\
&<\frac1{e^{\ln(y)}}-\frac1y\\
&=0.\end{align*}
Therefore, this means that $r>\frac{y-1}y$ by the Intermediate Value Theorem. Now, if $n=(x-1)\ln\left(\frac{x^2}{2x-1}\right)$, then we claim that $r<\frac{x-1}x$. We have
\begin{align*}
\left(\frac{x-1}x\right)^n-\frac{2x-1}{x^2}&=\frac1{\left(1+\frac1{x-1}\right)^n}-\frac{2x-1}{x^2}\\
&=\frac1{\left(1+\frac1{x-1}\right)^{(x-1)\ln\left(\frac{x^2}{2x-1}\right)}}-\frac{2x-1}{x^2}\\
&>\frac{2x-1}{x^2}-\frac{2x-1}{x^2}\\
&=0.\end{align*}
Therefore, by the Intermediate Value Theorem, $r<\frac{x-1}x$. Therefore, we have $x>y$. We claim that $m<2n$. Suppose that $m\geq2n$. Then, we have
\begin{align*}
x\ln(x)&>y\ln(y)\\
&=m\\
&\geq2n\\
&=(2x-2)\ln\left(\frac{x^2}{2x-1}\right)\\
&>(2x-2)\ln\left(\frac x2\right).\end{align*}
Therefore, we must have $x^x>\frac{x^{2x-2}}{2^{2x-2}}$, so $2^{2x-2}>x^{x-2}$. If $x>6$, then $2x-2<\frac52(x-2)$, which means that $x<2^{\frac52}=\sqrt{32}<6$. This means that $y<6$, so $m<6\cdot2=12$ and $n\leq\frac m2<6$. Since $a^n+a^2-1\mid a^m+a-1$, substituting $a=2$ gives $2^n+3\mid2^m+1$. Since $n<6$, this means that we must have $3\leq n\leq5$, so we must have either $11\mid2^m+1$, $19\mid2^m+1$, or $35\mid2^m+1$. If $11\mid2^m+1$, then $m\equiv5\pmod{10}$, so since $m<12$, we must have $m=5$ and $n=3$, which satisfies $m<2n$. If $n=4$, then $19\mid2^m+1$, which means $m\equiv9\pmod{18}$. Since $m<12$, we must have $m=9$. However, substituting $a=-2$ gives $19\mid-515$, which is not true. If $n=5$, then $7\mid2^m+1$, which is impossible.

Therefore, $m<2n$. Now, since $a^n+a^2-1\mid a^m+a-1$, we must have $a^n+a^2-1\mid a^m+a-1-a^{m-n}(a^n+a^2-1)=-a^{m-n+2}+a^{m-n}+a-1$. Therefore, we must have $m-n+2\geq n$, so $m\geq2n-2$. If $m=2n-2$, then $a^n+a^2-1\mid-a^n+a^{n-2}+a-1$, so $a^n+a^2-1\mid-a^n+a^{n-2}+a-1+a^n+a^2-1=a^{n-2}+a^2+a-2$, which is impossible since $n>n-2$ and $a^{n-2}+a^2+a-2\neq0$. If $m=2n-1$, then $a^n+a^2-1\mid-a^{n+1}+a^{n-1}+a-1$, so $a^n+a^2-1\mid-a^{n+1}+a^{n-1}+a-1+a(a^n+a^2-1)=a^{n-1}+a^3-1$. This is only possible when $n\leq3$, so since $n\geq3$, we must have $n=3$, so $m=5$, which works since $a^3+a^2-1\mid a^2+a^3-1$.

Therefore, the only pair of positive integers that works is $(m,n)=\boxed{(5,3)}$.

Here is a weird reason why $m<2n$ which I can't find above, given that we already know $x^n+x^2-1$ has to divide $x^{m-n+1}+x^{m-n}-1$.

Both of these polynomials are monotonic when $x>0$, and one divides the other, so it is impossible for them to have different signs for some $x>0$. Consider $x=\frac{1}{2}^{\frac{1}{n}}$. $\left(\frac{1}{2}^{\frac{1}{n}}\right)^n+\left(\frac{1}{2}^{\frac{1}{n}}\right)^2-1=\frac{1}{2}+\frac{1}{2}^{\frac{2}{n}}-1>0$ since $\frac{2}{n}<1$. On the other hand, $\left(\frac{1}{2}^{\frac{1}{n}}\right)^{m-n+1}+\left(\frac{1}{2}^{\frac{1}{n}}\right)^{m-n}-1=\left(\frac{1}{2}^{\frac{1}{n}}+1\right)\left(\frac{1}{2}^{\frac{m-n}{n}}\right)-1$. The first part of the product is $<2$ and since $\frac{m-n}{n}\ge 1$, the second part of the product is $\le \frac{1}{2}$. Thus this is $<0$, giving the desired contradiction.

Let $f(x)=x^n+x^2-1$ and $g(x)=x^m+x-1$ be polynomials in $\mathbb Z[x]$. We claim that $(m,n)=(5,3)$ works.
[b][color=#f00]Claim 1:[/color][/b] $f \mid g$ holds always.
[b][color=#f00]Proof:[/color][/b] For any $x \in \mathbb C$ we do polynomial division with quotient and residue in $\mathbb Z[x]$
$$g(x)=f(x)q(x)+r(x) \implies \frac{g(x)}{f(x)}=q(x)+\frac{r(x)}{f(x)}$$
From here note that $\text{deg} \; r<\text{deg} \; f$ so set $x$ to be a extremely large positive integer such that $-f(x)<r(x)<f(x)$ meaning that for infinite values $r$ is 0 so $r(x)=0$ holds always meaning that $f \mid g$ as desired.
[b][color=#f00]Claim 2:[/color][/b] $2n-1 \ge m$
[b][color=#f00]Proof:[/color][/b] From [b]Claim 1[/b] one has that all the roots of $f$ are also roots of $g$ now $f'(x)=nx^{n-1}+2x>0$ if $x>0$ hence $f$ is strictly increasing in $\mathbb R^+$, now $f(1)=1$ and $f\left(\frac{2}{3} \right)=\left(\frac{2}{3} \right)^n-\frac{5}{9}<0$ meaning that there exists a root of $f$ satisfying $\frac{2}{3}<z<1$.
Assume FTSOC that $m \ge 2n$ then by all said before and ineqs
$$(1-z^2)^2=z^{2n}\ge z^m=1-z \implies (1-z)(z^2+z+1+z) \ge 1 \implies 1-z^3+z-z^2 \ge 1 \implies 1 \ge z+z^2 \ge 3z-1 \implies \frac{2}{3} \ge z \; \text{contradicction!!}$$
Contradiction comes from assuming $m \ge 2n$ hence $2n-1 \ge m$ as desired.
[b][color=#f00]A nice finish:[/color][/b] Lets now work in integers and divisions
$$a^n+a^2-1 \mid a^m+a-1 \implies a^n+a^2-1 \mid a^{m-n+2}-a^{m-n}+1-a=(a-1)(a^{m-n}(a+1)-1) \implies a^n+a^2-1 \mid a^{m-n}(a+1)-1$$
From here as RHS has at most degree $n$ we see that both have to be equal and $m=2n-1$ hence $a^n+a^2-1=a^n+a^{n-1}-1$ which means $n=3$ and this means $m=5$.
Hence the pair $(m,n)=(5,3)$ is the unique pair that works, thus we are done :D

Did it back like 2 years ago? Very cool problem. Solved with [b][url=https://artofproblemsolving.com/community/user/543168]quirtt[/url][/b] and [b][url=https://artofproblemsolving.com/community/user/711847]Fafnir[/url][/b].

[i]Solution:[/i] We will show that $(m,n)=(5,3)$ is the only solution. One can see that it works as
\[a^5+a-1=(a^2-a+1)(a^3+a^2-1).\]
We will break the solution into two claims which will bound the values of $m$. We define \begin{align*} P(a) \overset{\text{def}}{=} a^m+a-1 \\ Q(a) \overset{\text{def}}{=} a^n+a^2-1 \end{align*} where $P,Q$ are polynomials.  We will now prove a well-known but extremely useful lemma which will take the problem from Number Theory to Algebra!

[b][color=blue]Lemma:[/b][/color] If two polynomials $f,g \in \mathbb{Z}[x]$ are such that $f(n) \mid g(n)$ for infinitely many integers $n$, then $f(x) \mid g(x)$.

[i]Proof:[/i] Due to size reasons, it is easy to see $\deg(g) > \deg(f)$. By the division algorithm over $\mathbb{Q}[x]$, write $g = fq + r$ for $f,q \in \mathbb{Q}[x]$ and $\deg(r) < \deg(f)$. One can clear out the denominators by multiplying by suitable integer, so we assume that all the polynomials are in $\mathbb{Z}[x]$. Just pick huge $k$ such that $f(k) \mid g(k)$. It would actually mean that $f(k) \mid r(k)$. But since $\deg(r) < \deg(f)$, we must have $r \equiv 0$ as desired. $\square$
-----------------------------------
So, we can actually apply the lemma to reduce the problem down to just two polynomials dividing each other!

[b][color=red]Claim:[/b][/color] $2n-1 \le m$.

[i]Proof:[/i] First of all notice that $\gcd(a+1,Q(a))=1$. The algebra for divisibility begins here \begin{align*} P(a)                   & \equiv 0 \pmod{Q(a)} \\ (a+1)P(a)              & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^2-a+ a^m+a-1 & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^m+a^2-1      & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^m-a^n        & \equiv 0 \pmod{Q(a)} \\ a^{m-n+1}+a^{m-n}-1    & \equiv 0 \pmod{Q(a)} \\ \end{align*} Bringing this into divisibility condition, we get \begin{align} a^n+a^2-1 \mid a^{m-n+1}+a^{m-n}-1 \label{1} \end{align} Due the bound on the degree, we get $m-n+1 \ge n$ which on simplifying gives $2n-1 \le m$ as desired. $\square$
-----------------------------------
The next is claim is a little harder to come up with and prove.

[b][color=red]Claim:[/b][/color] $m < 2n$.

[i]Proof:[/i] Assume for the sake of contradiction, that $m>2n$. By the [i]Intermediate Value Theorem[/i], there exists a real root of $Q$ which lies between 0 and 1. Let this root be $\alpha$. By the problem condition, $\alpha$ is also a root of $P$. Now $P(\alpha)$ and $Q(\alpha)$ gives the following \begin{align} \alpha^m + \alpha = 1 \\ \alpha^n + \alpha^2 = 1 \end{align} Now we compute $\alpha^m$ and $\alpha^{2n}$. \begin{align*} \alpha^m    & = 1- \alpha      \\ \alpha^{2n} & = (1-\alpha^2)^2 \\ \end{align*} By assumption, we must have $\alpha^m < \alpha^{2n}$. Plugging in the values, we get \begin{align*} 1-\alpha    & \le (1-\alpha^2)^2           \\ 1-\alpha    & \le (1-\alpha)^2(1+\alpha)^2 \\ 1           & \le (1-\alpha)(1+\alpha)^2   \\ \iff \alpha & \le \varphi \end{align*} where $\varphi$ is the golden ratio. We will now show that $\alpha > \varphi$ which would bring the desired contradiction.  It is easy to see that $P$ is an increasing function in the interval $[0,1]$. Note that \begin{align*} P(\varphi)=\varphi^m+\varphi-1 & < \varphi^2+\varphi-1 = 0 = P(\alpha) \\ \implies P(\varphi)            & < P(\alpha) \end{align*} and we can conclude that $\alpha > \varphi$ and we are done with the claim. $\square$
-----------------------------------
 Now, due to the bounds we just need to check the case for $m=2n-1$. Putting this in $(1)$ we get
\[a^n+a^2-1 \mid a^{n-1}-a^2.\]
This forces $n=3$ and we are done. $\blacksquare$

solved with hint to consider r, otherwise im very proud of myself :) :) :)

It's obvious that m>n; we want to find m,n s.t. $a^{m-n+2}-a^{m-n}-a+1\equiv a^m+a-1-a^{m-n}(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}$.
We claim that $m\le 2n$.
Assume FTSOC m>2n, and take 0<r<1 as the real number root of both polynomials. We have $$0=r^m+r-1\le r^{2n+1}+r-1=-(1-r)(r^4+r^3-r^2-r+1),r^4+r^3-r^2-r+1>r^2(r^2+r-1)>r^2(r^m+r-1)=0,1-r>0\implies 0\le RHS<0,$$ impossible.

Case 1. m=2n. We have $a(a^3-2a+1)\equiv a^{n+2}-a^n-a+1-(a^2-1)(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}\implies n=\{1,2,3\}$, none of which work.
Case 2. m=2n-1. We have $a^{n-1}+a^3-1\equiv a^{n+1}-a^{n-1}-a+1-a(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}$, and checking the possible n, only $\boxed{n=3,m=5}$ works.
Case 3. m=2n-2 has similar computation with degrees on n, and none of n work. Case 4 has $m\le 2n-3$, none of which work by looking at deg LHS<deg RHS except for a few n, and checking those don't work. Having exhausted all cases, we conclude the proof here.

[b]Remark.[/b] After enough of these problems, reducing the degree of polynomials by suitably subtracting multiples of the modulus becomes second nature; it's very natural to do so by euclidean algorithm, and once LHS|RHS with LHS deg<RHS deg except for a few numbers, the problem is solved.

The answer is $(m,n)=(5,3)$ only, which clearly works since $a^3+a^2-1 \mid a^5+a-1$ in the polynomial sense. We now prove this is the only one.

For fixed $(m,n)$, since $a^n+a^2-1$ is monic we can write
$$\frac{a^m+a-1}{a^n+a^2-1}=p(a)+\frac{q(a)}{a^n+a^2-1}$$
for some $p,q \in \mathbb{Z}[x]$ and $\deg q<n$. Then by considering sufficiently large $a$ we require $q \equiv 0$, so we should have $a^n+a^2-1 \mid a^m+a-1$ in the polynomial sense. Note that both of these polynomials have exactly one positive real root, so they must be the same. Call this common root $r$; clearly $r<1$.

[b][color=#00f]Claim:[/color][/b] $m<2n$.
[color=#00f]Proof:[/color] Suppose otherwise. We should then have
$$1=r^n+r^2=r^m+r<r^{2n}+r \implies r-r^2>r^n-r^{2mn}.$$
On the other hand, since $r^m<r$ but $r^n=1-r^2>1-r$, this is impossible, since $x-x^2$ increases as $x$ gets closer to $1/2$. $\blacksquare$

Now let $m=n+k$ with $0 \leq k \leq n-1$ (since clearly $m \geq n$). We should have
$$a^n+a^2-1 \mid a^k(a^n+a^2-1)-(a^{n+k}+a-1)=a^{k+2}-a^k-a+1=(a-1)(a^{k+1}+a^k-1).$$
Since $\gcd(a^n+a^2-1,a-1)=1$ (in the polynomial sense) it follows that $a^n+a^2-1 \mid a^{k+1}+a^k-1$. In general $a^{k+1}+a^k-1$ is a nonzero polynomial, so $k<n-1$ is impossible, hence $k=n-1$ and $a^n+a^2-1 \mid a^n+a^{n-1}-1 \implies a^n+a^2-1 \mid a^{n-1}-a^2$. The last polynomial has degree less than $n$, hence it must be zero, i.e. $n=3$. Thus we extract the claimed solution. $\blacksquare$

[b]Remark:[/b] Good problem to test "polynomial" intuition? To me considering $r$ felt "intuitive" but it's not super motivated in itself. I guess you start thinking about the roots after divisibility, and these are mostly hard to pin down except $r$.

The only answer is $(m, n) = (5, 3)$.

Obviously $m > n$. It's not hard to see that $x^n + x^2 - 1 \mid x^m + x - 1$. Let $r$ be a real root of $x^n + x^2 - 1$. Then $r$ is real root of $x^m + x - 1$. Thus $r^n + r^2 = r^m + r$, which is equivalent to $r(1 - r) = r^n(1 - r^{m-n})$.

If $m > 2n$, then $r(1 - r) = r^n(1 - r^{m-n}) > r^n(1 - r^{n}) \ge r(1 - r)$, a contradiction. Thus $m \le 2n$.

Now note that $x^n + x^2 - 1 \mid (x + 1)(x^m + x - 1) = x^{m+1} + x^m + x^2 - 1$, thus $x^n + x^2 - 1 \mid x^{m-n}(x + 1) - 1$. Since $m - n + 1 \le n + 1$, thus $x^{m-n}(x + 1) - 1 = (x + 1)(x^n + x^2 - 1)$ or $x^{m-n}(x + 1) - 1 = x^n + x^2 - 1$. In the first case we have $x^{n+1} + x^n + x^3 + x^2 - x - 1 = x^{n} + x^2 - 1$, which is impossible. In the second case we have $x^n + x^2 - 1 = x^n + x^{n-1} - 1$, thus $n = 3$. Thus $m = 5$. Therefore $(m, n) = (5, 3)$ is only answer. $\blacksquare$

Note that these are both polynomials, so if for infinitely many $a$, $a^n+a^2-1\mid a^m+a-1$ then the polynomial $a^n+a^2-1$ divides the polynomial $a^m+a-1$. Note that $P(x) \mid Q(x)$ implies $\text{deg} P\le \text{deg} Q$. In particular, $m\ge n$.

Since $a^n + a^2 - 1\mid a^m+a-1$, which implies $a^n+a^2-1|(a+1)(a^m+a-1)-(a^n+a^2-1)$. This becomes $a^n+a^2-1\mid a^{m+1}+a^{m}-a^n$. Note that $\gcd(a^n+a^2-1,a)=\gcd(a^2-1,a)=1$ so $a^n+a^2-1\mid a^{m-n+1}+a^{m-n}-1$. This implies $m\ge 2n-1$. This also means $m-n\ge n-1\ge 2$.

By IVT and taking the derivative, there exists a unique real $0<r<1$ for which $r^n+r^2-1=0$. We must have $r^m+r-1=0$. Setting them equal, we get
\[r=\frac{r^n(1-r^{m-n})}{1-r}=\frac{(1-r^2)(1-r^{m-n})}{1-r}=(1+r)(1-r^{m-n})\] and this becomes $r^{m-n}(r+1)=1=r^n+r^2\ge r^n+r^{m-n}$ which implies $r^{m-n+1}\ge r^n\implies m-n+1\le n$ so $m\le 2n-1$. Note that equality is dependent on $m-n=2$, so we need $(m,n)=(5,3)$ which works.

[b][color=#f00]Claim.[/b][/color] [Main Claim] $m \leq 2n$.

[i]Proof.[/i] Let $r$ be a root of $a^n + a^2 - 1 = 0$, such that $r^n + r^2 = r^m + r = 1$. Suppose for the sake of contradiction that $m > 2n$. Then,
\begin{align*}
r^m + r &< r^{2n} + r \\
&= (r^n - r^{n+2}) + r \\
&= 1 - 2r^2 + r^3 + r \\
0 &< r^4 + r - 2r^2 \\
&=r(r-1)(r^2+r-1).
\end{align*}
On the other hand, $r^2 + r > 1$ and $r - 1 < 0$, so this is a clear contradiction. $\blacksquare$

Now, by Euclidean algorithm we have $$a^n + a^2 - 1 \mid (a-1)(a^{m-n+1}+a^{m-n} - 1).$$ In particular, as $a^n + a^2 - 1 \equiv 1 \pmod {a-1}$, we have $$a^n + a^2 - 1 \mid a^{m-n+1} + a^{m-n} + 1.$$ If $m \leq 2n-2$, then the RHS is clearly smaller and $a^{m-n+1}+a^{m-n} + 1 = 0$, which is clearly impossible.

If $m \in \{2n-1, 2n\}$, perform one more step to get $$a^n + a^2 - 1 \mid a^{m-n} + a^{m-2n+1}-a^{m+3-2n} - 1.$$ As $m-2n+1 \leq 1$, the RHS is smaller than the LHS now, so it must uniquely equal zero. When $m = 2n-1$, we get $a^{n-1} = a^2$, giving the solution pair $(3, 5)$. When $m = 2n$, we get $a^n + a - a^3 - 1=0$, which does not work for any value of $n$. Thus, $(3, 5)$ is the only solution.


Circles A and B are externally tangent to each other and to line [i]t[/i]. Find the area in between the 2 circles and its external tangent if the sum of the radii of the two circles is 12 and the radius of circle A is 3 times that of circle B.

[hide="Hint"]The sum of the radii is 12 and the radius can be expressed as $4x$ and $x$. So $4x+x=12$. Which gets us fractional radii.Then find the length of the external tangent. Then find the area of a triangle(if you found the external tangent) and the rectangle. Then use central angle measures to help you subtract to find the area. [/hide]
Are you sure you wrote this right?

[hide]

From simple algebra, I got the radii to be $\frac{12}{5}$ and $\frac{48}{5}$. The radii are perpendicular to the tangent, so the line beween the two points, the two radii, and the tangent make a quadlerateral which can be divided into a parallelogram and a right triangle (one leg of the triangle is the tangent, and the hypotnuse is a side of the parallelogram). You can find the area of these figures, and then subtract the area of the sectors to get the area we're looking for. However, my calculations gave the angles to be really weird; according to my calculator one of them is $26.56505118...$ degrees

which you could use with a calculator but the answer probably won't be very clean.

[/hide]

[hide]
Find the area (tangent(A, line-t), tangent(B, line-t), B, A) by dropping a parallel line from the tangent.

$r+R = 12$
$R = 3r$
$4r = 12 \implies r = 3$
$R = 9$

Whole area is $(0.5 * 6 * \sqrt{12^{2}-6^{2}})+(3 * \sqrt{12^{2}-6^{2}}) = 6\sqrt{108}$

Central angle of A is $60^\circ$ from trigonometry, central angle of B is $120^\circ$.

Subtracting the appropriate areas gives: $\boxed{36\sqrt{3}-\frac{33\pi}{2}}$
[/hide]

Ah for some reason I thought the problem said one radius was FOUR times the other... that would explain why my answer was so messy.

I think it did say 4 times (I remember reading that as well..it mustve been a typo by shark..)

nice solution javaman


geometry
geometric transformation
dilation

If Henry's law is $ P \equal{} K_hc$ where $ c$ is the concentration then how does it get converted into $ P \equal{} K_h\chi$ where $ \chi$ is mole fraction. Isnt it completely different?

Henry's Law states that $ P_i \equal{} K \cdot x_i^L$, where $ P_i$ is the partial pressure of the solute i in the vapor above an ideally dilute solution, $ K$ is a constant for i, and $ x_i^L$ is the mole faction of i in the solution.

[quote="Conjurer"]In many texts, mole fraction is replaced by concentration of dissolved solute.We have diff. value of K_h then, but how can we do like that.

wiki says the following:
[quote]
For a dilute solution, the concentration of the solute is approximately proportional to its mole fraction x, and Henry's law can be written as:

    $ p \equal{} k_{\rm H}\,x$ [/quote][/quote]

Henry's law is typically expressed as: $ P_i \equal{} K_H c$.

The concentration, $ c$ is the molar density of the solute, i.e. $ c \equal{}\frac{N_i}{V} \equal{} n_i$.

The mole fraction is defined as: $ x\stackrel{\triangle}\equal{}\frac{n_i}{n}$ , where $ n$ is the molar density of the solution.

If $ c \equal{} n_i$ then $ c \equal{} x n$.

Substituting this into the expression for Henry's law: $ P_i \equal{} K_H n x \equal{} K_{H}^{\prime} x$.

$ K_H$ is not actually constant but is dependent on the solvent used; therefore it is alright to replace it with $ K_{H}^{\prime}$.


Let $\delta (\mathcal{A})$ denote the Schnirelmann density of a set. Is it true that $\delta (\mathcal{A}) > 0$ implies $\mathcal{A}$ contains infinitely many prime ? I believe this is trivially false, but since I'm just beggining to learn additive number theory, I would like to hear some solid arguments. Thanks ;)

Thanks !
And glumps ;), I think I'm tired, the question was silly ;)

[quote="ZetaX"]Take the set of all nonprimes (the result follows e.g. from the prime number theorem).[/quote]Actually, the S's density of nonprimes is $\frac{1}{3}$. It's achieved only ones, for n=3 (the prime number theorem is an overkill (:)) ).


analytic geometry
graphing lines
slope
trigonometry
ratio
Pythagorean Theorem
geometry

Hi all, i've been having trouble with this problem it reads

" an equation of a line that runs through the point (-1,2) with the slope of 2/3."

A. 3x-2y=-8
B. 3x-2y=7
C. 2x-3y=-8
D. 2x-3y=8
E. 3x-2y=7

i went through each one at least two times and solved for Y, to get the euqation of the line y=mx+b
now the only way i could get a 2/3 slope was to use option D and C because in the other ones when slving for Y the numerator becomes 3. now this is what i thought was right and i thought the answer must be either C or D but none of them give me a line that goes through the point (-1,2). i feel like i should be able to solve this and it is really frustrating me. thanks

[quote]but none of them give me a line that goes through the point (-1,2)[/quote]

Plug in $ (\minus{}1,2)$ into C. What's the problem?

lol is that all you do? just have to see if the calculated number and the original match? wow ok thanks

That's how you'll do it if you have multiple choice. If there weren't solutions, then you'd do it like this:

$ y \equal{} kx \plus{} n$
$ y \equal{} \frac {2}{3}x \plus{} n$

And now, you need to find $ n$, and taking into account the fact that our equation is true for $ ( \minus{} 1,2)$, plug it in.

Now you have $ 2 \equal{} \minus{} \frac {2}{3} \plus{} n$, so $ n \equal{} \frac {8}{3}$, so the equation is

$ y \equal{} \frac {2}{3}x \plus{} \frac {8}{3}$ or equivalently $ \minus{}3y \plus{} 2x \equal{} \minus{}8$ qed.

Well, two advices: 1. for a new problem, open a new thread. 2. this should probably go to high school basics, or something like that, but anyways:

recall that $ \sin^2\phi \plus{} \cos^2\phi \equal{} 1$, for all $ \phi$. Now the answer should be obvious.

lol maybe this was basic for you man, but we didn't covered this in pre calc which is what i took my last  year at high school..ok thanks a lot for the help

don't know what you mean by sqr, but it doesn't make a difference.  Anyway $ a^2\plus{}b^2\equal{}c^2$ (pythagorean theorem) implies $ (a/c)^2\plus{}(b/c)^2 \equal{} 1$.  Of these two ratios, one must be a sine and one a cosine, hence $ \sin^2 \theta \plus{} \cos^2 \theta \equal{} 1$.  Some other nice identities, which you probably didn't cover:

pythagorean
$ 1\plus{}\tan^2 \theta \equal{} \sec^2 \theta$
$ 1\plus{}\cot^2 \theta \equal{} \csc^2 \theta$

reciprocal
$ \cot \theta \equal{} \frac{1}{\tan \theta}$
$ \csc \theta \equal{} \frac{1}{\sin \theta}$
$ \sec \theta \equal{} \frac{1}{\cos \theta}$

cofunctions:
$ \sin(\pi/2\minus{}\theta) \equal{} \cos \theta$
$ \tan(\pi/2\minus{}\theta) \equal{} \cot \theta$
$ \sec(\pi/2\minus{}\theta) \equal{} \csc \theta$

Even/odd
If $ f$ is sine, tangent, cosecant or cotangent, $ f(\minus{}x)\equal{}\minus{}f(x)$
If $ f$ is cosine or secant, $ f(\minus{}x)\equal{}f(x)$

quotient:
$ \tan \theta \equal{} \frac{\sin \theta}{\cos \theta}$
$ \cot \theta \equal{} \frac{\cos \theta}{\sin \theta}$

Some more complicated:
Sum and difference formulas
$ \sin(a \pm b) \equal{} \sin a \cos b \pm \cos a \sin b$
$ \cos(a \pm b) \equal{} \cos a \cos b \mp \sin a \sin b$
$ \tan(a \pm b) \equal{} \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$

From which we get double angle formulas if $ a\equal{}b$.
$ \sin(2a) \equal{} 2\sin a \cos a$
$ \cos(2a) \equal{} \cos^2a \minus{} \sin^2a \equal{} 1\minus{}2\sin^2a \equal{} 2\cos^2a\minus{}1$
$ \tan(2a) \equal{} \frac{2\tan a}{1\minus{}\tan^2 a}$

Power reducing formulas come from the double angle formulas for cosine, and half angle formulas are basically the square roots of these formulas, with a substitution of the form $ a\equal{}b/2$ (tangent's half angle formula can be simplified several ways to remove the radical).  After that there's product to sum and sum to product formulas, which can be verified by (and difficultly derived from) the sum and difference formulas.

[quote="kevinman45"]wow thanks a lot, do you know how to even start this one out?

"for all positive real numbers X, 1-sin^2(Sqr X)="

1-sinx
cosx
0
cos^2(sqr x)
1

those are the choices.[/quote]
I have a feeling that by "Sqr X", you mean $ \sqrt{x}$.
hsiljak and facis gave you some identities, and one of them comes particularly in handy here...
[hide="hint"]$ \sin^2 \theta \plus{} \cos^2 \theta \equal{} 1$.
[hide="Even more hint"]$ \cos^2 \theta \equal{} 1 \minus{} \sin^2 \theta$[/hide][/hide]


inequalities
geometry
angle bisector
triangle inequality
inequalities solved

In  any  triangle  ABC  prove  that
ma+mb+mc+min(a,b,c)<=la+lb+lc+max(a,b,c)

where  ma,mb,mc  the  medians  and  la,lb,lc  the  bisectors  of  the  triangle  ABC.
THANKS!

Let AA2 the interior angle bisector of angle A and AA1 the median of triangle ABC. Applying the triangle inequality for AA1A2 we obtain:

    |ma - la| \leq |A1A2|

                or

    ma - la \leq |a/2 - ca/b+c| = a|b-c|/2(b+c) \leq |b-c|/2.
Therefore:

     \sum (ma - la) \leq (|a-b| + |b-c| + |c-a|)/2.
Following the identity

    |a-b| + |b-c| + |c-a| = 2max(a,b,c) - 2min(a,b,c)
we obtain our inequality.
    The same way we can obtain some inequalities:

ma + mb + mc + 2 min(a,b,c) \leq sa + sb + sc + 2max(a,b,c), where sa, sb, sc are the simedians.

ga + gb + gc + 2max(a,b,c) \geq na + nb + nc + 2min(a,b,c), where ga, gb, gc is Gergonne`s line and n<sub>a</sub>, n<sub>b</sub>, n<sub>c</sub> is Nagel`s line.


number theory
greatest common divisor
algebra
polynomial
Rational Root Theorem

Prove that, for $ n \ge 2$, $ n^{\frac{1}{n}}$ is irrational.

[quote="AndrewTom"]Prove that, for $ n \ge 2$, $ n^{\frac {1}{n}}$ is irrational.[/quote]

I'm going to split this up into cases.

Case 1
---------

Suppose n is a prime number.

Let $ n \equal{} p$ and let $ n \equal{} \frac ab$ where $ a$ and $ b$ are positive integers,
and we suppose $ gcd($a, b$ ) \equal{} 1$.
That is, the fraction is as reduced as possible.

Then $ p^{\frac1p} \equal{} \frac ab$

$ (p^{\frac1p})^p \equal{} (\frac ab)^p$

$ p \equal{} \frac {a^p}{b^p}$ . . . (This is $ n \equal{} \frac {a^n}{b^n}$.)

$ pb^p \equal{} a^p$

Then $ p|a^p$

$ p|a^p$ means that $ p|a$ because $ p$ is a prime number.

Let $ a \equal{} kp$, for some integer $ k$

Then $ (a)^p \equal{} (kp)^p$

And $ a^p \equal{} k^pp^p$

Because $ a^p \equal{} pb^p$, then

$ pb^p \equal{} k^pp^p$ because they are both equal to $ a^p$

Divide both sides by p:

$ b^p \equal{} k^pp^{p \minus{} 1}$

Then $ p^{p \minus{} 1}|b^p$

$ p \minus{} 1 \ge 1$ because $ p \ge 2$

So $ p^{p \minus{} 1}|b^p$ means that $ p|b^p$

$ p|b^p$ means that $ p|b$ because $ p$ is a prime number

This is a contradiction because $ \frac ab$ was assumed to be a fraction
in reduced form.

---------------------------------------------------------------------------------------

Case 2
---------

Suppose $ n$ is a composite number.

Let $ n \equal{} mp^r$, where $ p$ is a prime number, $ r$ is the
total number of factors of the specific prime, $ p$,
and $ m$ and $ r$ are positive integers.

If $ r \ge 2$, then $ m$ is an appropriate positive integer.
If $ r \equal{} 1$, then $ m$ is an appropriate positive integer that is
greater than or equal to 2.

Then $ gcd(p, m) \equal{} 1$

As in case 1, $ n \equal{} \frac {a^n}{b^n}$

So $ mp^r \equal{} \frac {a^{mp^r}}{b^{mp^r}}$

Then $ mp^rb^{mp^r} \equal{} a^{mp^r}$ . . . (*)

$ p^r|a^{mp^r}$

$ p|a^{mp^r}$, and it means that $ p|a$

Let $ a \equal{} kp$, for some integer $ k$

$ (a)^{mp^r} \equal{} (kp)^{mp^r}$

$ a^{mp^r} \equal{} k^{mp^r}p^{mp^r}$

Then $ mp^rb^{mp^r} \equal{} k^{mp^r}p^{mp^r}$ because of (*)

Divide both sides by $ p^r$:

$ mb^{mp^r} \equal{} k^{mp^r}p^{(mp^r \minus{} r)}$

$ (mp^r \minus{} r) \ge 1$ because $ p \ge 2$

$ p^{(mp^r \minus{} r)}$ does not divide $ m$ because $ gcd(p, m) \equal{} 1$

So $ p^{(mp^r \minus{} r)}|b^{(mp \minus{} r)}$

Then $ p|b^{(mp \minus{} r)}$

And $ p|b$ because $ p$ is a prime number.

This is a contradiction because $ \frac ab$ was assumed to be a fraction
in reduced form.

==========================

Therefore, $ n^{\frac 1n}$ is irrational for $ n \ge 2$.

it's a standard result. by the rational root theorem, if $ x^n\minus{}n\equal{}0$ has a rational root, $ \frac{p}{q}$ then $ q|1$ and $ p|n$

So $ q\equal{}1$ and $ p^n\minus{}n\equal{}0$, but clearly $ 1^n<n<2^n$ so $ p^n\minus{}n\equal{}0$ has no integer solutions and $ n^{\frac{1}{n}}$ is irrational

[quote="TRAN THAI HUNG"][quote="AndrewTom"]Prove that, for $ n \ge 2$, $ n^{\frac {1}{n}}$ is irrational.[/quote]
n is an integer or a rational? :)[/quote]

The variable, n, is an integer, but it should be stated as such in the
problem, regardless that "n" is typically used for integers.


calculus
AMC
AMC 12
AMC 10
AIME
geometry
3D geometry

does your school give both tests or just one?

[color=cyan]Our school gives both, but we are a bit special.  That is, more students from Stuyvesant take the test than anywhere else in the country.  1,500 or more last year took the A exam, about half of the school.  Pretty cool, huh?  Most of that is because it's required for high-level math classes (definitely multi and diffeq, I think BC calculus, and maybe AB) and math team (which has more than 200 people, mostly freshmen).  Anyhow, I believe more than 300 people take the B exam, usually.

My old school which I left after 8th grade but which goes K-12 gives the test on the B day because they're on vacation on the A day.  But not many people take it.[/color]

At my high school we took both, but we only ordered 40 AMC12s and 60 AMC10s for the A and 10 of each for the B, so many people who wanted to take them couldn't, which was a shame.

At hunter, they offer both A and B as well

you go to hunter quantum?  i thought you went to Stuyvesant.  How do you, JBL, tetratomato all seem to know each other?

I wish my school offered both, maybe this year we can convince them to do so.  If you take both A & B, do you get to take both AIME's too, or just one?  thanks.

I went to hunter, and transferred to stuyvesant.
I, Joel, and teratomato know each other cuz we all go to stuy (well, Joel just graduated), and teratomato transferred from hunter as well

No, each student may only take one of the AIME's.

We only get to take one AMC of a kind. You could take both 10 and 12, though, I think.

sigh. i wish my school gave both. do we have to take them through our school? i mean, i think NCSSM might give both, so could I go there and take the B test or must i take through my school?

[quote]How come you can do only 1 AIME?[/quote]

The AMC guys said so.

i know that some kids whose schools dont offer any AMC go to other schools who do.  I know someone who did, but for some reason his score didnt count... but i think it should count even if you dont do it at your school.

[color=cyan]You can only take one AIME because the reason they offer 2 AMCs is due to the fact that many schools are on vacation when they usually offer it and thus large groups of people were unable to do it.  This is not the case with the AIME.  The second AIME is just for the (very small) section of people who wouldn't be able to take the normal AIME.[/color]

[quote="Naismith"]Now that the AMC 10 has a guaranteed cutoff score of 120 that reasoning may no longer apply.
[/quote]

I'm not familiar with this rule.  Is it new for 2004?

I'm sure I saw a reference to automatically being able to qualify for the AIME by scoring 120 or above on the AMC 10 (starting in 2004) somewhere a few months ago, but I'll be darned if I can track that reference down now.  Maybe it was in the 2003 Summary of Results booklet; my copy is not on the shelf where it should be.  I suppose this should be considered "unconfirmed" for now.

earlier i was corrected on a phrase when i mentioned there was no amc 12 and backup amc 12. im sorry if i was misunderstood. i meant that unlike the aime, both test dates are equally important, ie there is no designated backup exam, a and b are essentially equal, as opposed to aime and alternate aime. i believe that one of the reasons there are two dates for amc10/12 was that students in 10th and below should get a chance to take both the amc 10 and the amc 12, and because of this, the juniors/seniors can take amc 12 twice. also, i disagree with fierytycoon about getting to school on the aime day. what if you are truly sick, or seriously injured. i cant see a person in the hospital gettign to the school no matter how dedicated they are. the existence of 2 aimes is necessary i think.

What about two USAMOs by the same reasoning?

Tetrahedr0n - now I see what you were saying about "no backup AMC" - your explanation is very clear.

Also - very good point about being in the hospital.  I would add to that other unexpected events like school closings due to flooding, power outages, etc.

Yes, those incidents do occur, but on a very small scale. A very very small percent of schools get flooded on the day of the AIME. And yeah, why not have the same policy for the USAMO then?

You'd have to ask the "AMC folks" as to why no dual-USAMO, but I can offer some guesses:

1) There are many fewer students taking the USAMO, so P(a student invited to take the USAMO has a problem with a particular date) is much smaller;

2) The time of the USAMO (early May) is not a time when any school in the US generally takes a holiday - vs. both the AMC (mid-Feb. holidays in several states) and AIME (among all the various spring break times).

3) It is much harder to create USAMO tests of equivalent difficulty for such a small test population.

[quote]3) It is much harder to create USAMO tests of equivalent difficulty for such a small test population.[/quote]

Equally hard to create AIME tests of equivalent difficulty.

[quote="ComplexZeta"]It's somewhat less important to create two AIMEs of equal difficulty as compared to the USAMO. The USAMO is used to decide the IMO team, so they really want to make sure that the top people are USAMO winners. If one test is significantly more difficult than the other (assuming they have two), a top student who takes the more difficult USAMO might not make top twelve.[/quote]

Wait...doesn't the MOSP test determine the IMO team?  Also, wouldn't a top student still make the top 12 if he or she takes the "harder" USAMO since it's going to be all relative with whoever takes which tests.

-interesting_move

going to moP doesnt mean you go to imo. its just a prep program tingy

i think tst and usamo are averaged together to determine the imo team. therefore a harder usamo 2 would still be very detrimental to some students.

Dear Readers of the AMC Forum --

At the suggestion of the creators of this forum, I am going to try to post as I am able,
My intent is to write to squash rumors and misinformation.

"Naismith" is basically correct in his information. Please re-read his posts, I have no need to repeat what he said.

Last year the statstics, including mean, variance, and cumulative distributions for scores for the AIMEI and the AIME II were about as close as any two distributions that I have ever seen.   It appears that after several years of experience the AMIE committtee has the hang of creating tests which are as nearly equal in difficulty, as measured by the actual total perfromance statistics, as can be humanly expected.  I commend them, and hope that it turns that way again this year!

The AMC 10/12 committee seems to be moving in the same direction.  I have great confidence in their ability to calibrate the contests at the correct level and make them equal.  In the grand statistics, I can find no overall advantage to one contest over another.

Note, your personal experience may differ.

Is the rumor that MOP will be slightly expanded this year true? Will there be a USAMO central event? thanks


Let $a,b,c \in \mathbb{R}^{+}$. Prove that
\[ \frac{b^2+c^2}{a^2+bc}+ \frac{a^2+c^2}{b^2+ca} + \frac{a^2+b^2}{c^2+ab } \geq 3  \]. When does equality hold?

This problem is easy .
 By AM-GM : $  bc \leq \frac{b^2+c^2}{2} $
  The ineq change into $ \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} \geq \frac{3}{2} $ , where $ x=b^2+c^2 $

  Could you please tell me what Indian IMOTC is ?

Indian IMOTC refers to the Indian TST for the International Mathematical Olympiad. TST ---  Training cum selection test. We call'em IMOTC's here!!!

we have $b^2+c^2\geq 2bc .$
 hence $A\geq \sum \dfrac{a^2+b^2}{a^2+b^2+2c^2 }$
  let $a^2+b^2=x ;p c^2+b^2=y; a^2+c^2=z$
 we have proof solution

Also not new... http://www.mathlinks.ro/Forum/viewtopic.php?t=22664

  darij


algebra
polynomial
induction
algebra solved

Let $P:\mathbb{C}\rightarrow\mathbb{C}$ be a polynomial of degree [i]at most[/i] $n$ such that, for $k=0,1,2,\ldots, n$,

$P(k)=2^k$

Find $P(n+1)$

This can be solved by the method of finite differences.

[hide]Define $P_m(k) = P_{m-1}(k+1)-P_{m-1}(k)$ for $n - m \ge k \ge m \ge 1$ where $P_0(k) = 2^k$. We use induction to show that $P_m(k) = 2^k$ for $m \ge k \ge 0$. We have the base $m=0$, and the inductive step from $m$ to $m+1$ follows directly from algebra. We find that $P_m(k)$ is non-constant for $m < n$, which together with the givens implies that $P$ is an $n$th degree polynomial. Then

$P(k+1) = \sum_{m=0}^{n} P_m(n-m) = \sum_{m=0}^{n} 2^{n-m} = 2^{n+1} - 1$.[/hide]

Yes.
Or you may use that $P(x) = \sum_{i=0}^n C_x^i$,
where $C_x^i = \frac {x(x-1) \cdots (x-i+1)} {i!}$

Pierre.

This is quite the same way as Mildorf... ;)

Pierre.

Or, just use Lagrange interpolation polynomial. The problem is IMO shortlisted, I think.


calculus
integration
trigonometry
calculus computations

First, we state without proof the following lemma.

$ \frac {d}{dx}\left[\tan^{ \minus{} 1} \frac {P(x)}{Q(x)}\right] \equal{} \frac {P'Q \minus{} PQ'}{P^2 \plus{} Q^2}$.

Next, we note that

$ x^6 \plus{} 14x^3 \minus{} 1 \equal{} (x^3)^2 \plus{} 14x^3 \plus{} 49 \minus{} 50 \equal{} (x^3 \plus{} 7)^2 \minus{} (5\sqrt {2})^2$
$ \equal{} (x^3 \plus{} 7 \minus{} 5\sqrt {2})(x^3 \plus{} 7 \plus{} 5\sqrt {2})$.

But since $ 7 \pm 5\sqrt {2} \equal{} (\sqrt {2} \pm 1)^3$, we find

$ x^6 \plus{} 14x^3 \plus{} 1 \equal{} (x \plus{} \alpha)(x^2 \minus{} \alpha x \plus{} \alpha^2)(x \plus{} \beta)(x \minus{} \beta x \plus{} \beta^2)$,

where $ \alpha \equal{} 1 \plus{} \sqrt {2}$, $ \beta \equal{} 1 \minus{} \sqrt {2}$.

Since $ \alpha \plus{} \beta \equal{} 2$, $ \alpha\beta \equal{} \minus{} 1$, we find $ (x \plus{} \alpha)(x \plus{} \beta) \equal{} x^2 \plus{} 2x \minus{} 1$, so

$ \frac {(x^2 \plus{} 1)(x^2 \plus{} 2x \minus{} 1)}{x^6 \plus{} 14x^3 \minus{} 1} \equal{} \frac {x^2 \plus{} 1}{(x^2 \minus{} \alpha x \plus{} \alpha^2)(x^2 \minus{} \beta x \plus{} \beta^2)}$.

From here, it is a tedious exercise in partial fraction decomposition to evaluate the integral.  Alternatively, we can write the denominator as

$ x^4 \minus{} (\alpha \plus{} \beta) x^3 \plus{} ((\alpha \plus{} \beta)^2 \minus{} \alpha\beta)x^2 \minus{} (\alpha\beta)(\alpha \plus{} \beta)x \plus{} (\alpha\beta)^2 \equal{} x^4 \minus{} 2x^3 \plus{} 5x^2 \plus{} 2x \plus{} 1$
$ \equal{} (x^4 \minus{} 2x^3 \minus{} x^2 \plus{} 2x \plus{} 1) \plus{} 6x^2 \equal{} (x^2 \minus{} x \minus{} 1)^2 \plus{} (x\sqrt {6})^2$.

Letting $ P(x) \equal{} x\sqrt {6}, Q(x) \equal{} x^2 \minus{} x \minus{} 1$, it follows $ P'(x) \equal{} \sqrt {6}$, $ Q'(x) \equal{} 2x \minus{} 1$, and

$ \frac {P'Q \minus{} PQ'}{P^2 \plus{} Q^2} \equal{} \frac {\sqrt {6}(x^2 \minus{} x \minus{} 1) \minus{} (x\sqrt {6})(2x \minus{} 1)}{x^4 \minus{} 2x^3 \plus{} 5x^2 \plus{} 2x \plus{} 1} \equal{} \frac { \minus{} \sqrt {6}(x^2 \plus{} 1)}{x^4 \minus{} 2x^3 \plus{} 5x^2 \plus{} 2x \plus{} 1}$.

Therefore

$ \int \frac {x^2 \plus{} 1}{x^4 \minus{} 2x^3 \plus{} 5x^2 \plus{} 2x \plus{} 1} \, dx \equal{} \frac {1}{\sqrt {6}} \tan^{ \minus{} 1}\frac {x\sqrt {6}}{1 \plus{} x \minus{} x^2}$.

The definite integral is a bit tricky to evaluate since it turns out the antiderivative is undefined at the lower limit $ x \equal{} (1 \plus{} \sqrt {5})/2$.  This is resolved by taking the right-hand limit.  The remaining calculation is boring and so is left as an exercise.  However, it should be noted that

$ r \equal{} \frac {1 \plus{} \sqrt {2} \plus{} \sqrt {7 \plus{} 2\sqrt {2}}}{2} \equal{} \frac {\alpha \plus{} \sqrt {\alpha^2 \plus{} 4}}{2}$,

and so this value satisfies $ r^2 \minus{} \alpha r \minus{} 1 \equal{} 0$, from which it follows that $ 1 \plus{} r \minus{} r^2 \equal{} \minus{} \sqrt {2}r$ and $ \frac {\sqrt {6}r}{1 \plus{} r \minus{} r^2} \equal{} \minus{} \sqrt {3}$.

The answer is $ \frac {\pi}{6\sqrt {6}}$.

[quote="nguyenvuthanhha"][i]Evaluate

$ \mathcal{I} \ \equal{} \ \large\int_{\frac{1\plus{} \sqrt{5}}{2}}^{\frac{1 \plus{} \sqrt{2} \plus{} \sqrt{7 \plus{} 2 \sqrt{2}} }{2}} \frac{(x^2 \plus{}1)( x^2 \plus{} 2x \minus{}1)}{x^6 \plus{} 14x^3 \minus{}1} \ \ dx$[/i][/quote]

$ \frac{(x^2 +1 )(x^2 +2x-1)}{x^6 + 14x^3 -1} = \frac{(1 + \frac{1}{x^2})(x+ 2 - \frac{1}{x}) }{ x^3 +14 - \frac{1}{x^3}} =  \frac{(1 + \frac{1}{x^2})(x+ 2 - \frac{1}{x}) }{ \left(x- \frac{1}{x} \right) \left( \left(x- \frac{1}{x} \right)^2 + 3 \right) +14 } $

  So ; the key to a very short solution is the denotation : $ t = x - \frac{1}{x}$

Then, as ${\rm d}t = \left(1 + \frac{1}{x^2}\right) {\rm d}x$, we can write

$\frac{(1+\frac{1}{x^{2}})(x+2-\frac{1}{x}) }{\left(x-\frac{1}{x}\right)\left(\left(x-\frac{1}{x}\right)^{2}+3\right)+14 } {\rm d}x = \frac{t + 2}{t\left(t^2 + 3\right) + 14} {\rm d}t = \frac{{\rm d}t}{t^2 - 2t + 7}$.

Using completion of the square we see that $t^2 - 2t + 7 = (t-1)^2 + 6$ hence the indefinite integral is equal to

$\frac{1}{\sqrt{6}} \arctan\left(\frac{t-1}{\sqrt{6}}\right)$.


inequalities
algebra
function
domain
inequalities proposed

If $ n\in \mathbb{N}^*\backslash \{1\}$ show that :

$ \left(1\plus{}\frac{1}{n}\right)^{2n\plus{}1}\le \left(\frac{n(n\plus{}2)(n^2\plus{}1)}{n^4}\right)^n\cdot \left(1\plus{}\frac{1}{n\plus{}1}\right)$

Why is $ 1$ not in the domain? And why did you write $ \frac {n}{n^4}$? (I will use $ x$ instead of $ n$) But if the statement is correct, then the inequality is equivalent to $ \frac {(x^3 \plus{} 2x^2 \plus{} x)^ {x \plus{} 1}}{x^{3x} (x \plus{} 1)(x^2 \plus{} 1) } \left(\left( 1 \plus{} \frac {2} {x^3 \plus{} 2x^2 \plus{} x} \right) ^{x \plus{} 1} \minus{} \left(1 \plus{} \frac {1}{x^2}\right) \right)\ge 0$. From Bernoulli we get $ \left( 1 \plus{} \frac {2} {x^3 \plus{} 2x^2 \plus{} x} \right) ^{x \plus{} 1} \ge 1 \plus{} \frac {2(x \plus{} 1)} {x^3 \plus{} 2x^2 \plus{} x} \equal{} 1 \plus{} \frac {2} {x(x \plus{} 1)} \ge 1 \plus{} \frac {1}{x^2}$ for $ x \ge 1$.

1 is not in the domain because in my solution i divide the inequality by $ n^2\minus{}1$ .

@alex2008: would you show me your solution?^^
My solution shows that your inequality holds for real numbers greater than or equal one. But what is the minimum value $ x_0$ for which the inequality is valid?

We have that :

$ \frac{n(n\plus{}2)(n^2\plus{}1)}{n^4}\equal{}\frac{n^2\minus{}1\plus{}n\plus{}1\plus{}n\minus{}1\plus{}1}{n^2\minus{}1}\cdot \frac{(n^2\minus{}1)(n^2\plus{}1)}{n^4}$

$ \equal{}\left(1\plus{}\frac{1}{n\minus{}1}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n^2\minus{}1}\right)$$ \cdot \frac{n^4\minus{}1}{n^4}\equal{}\left(1\plus{}\frac{1}{n\plus{}1}\right)\left(1\plus{}\frac{1}{n\minus{}1}\right)\left(1\minus{}\frac{1}{n^4}\right)$

So $ RHS\equal{}\left(1\plus{}\frac{1}{n\minus{}1}\right)^n\cdot \left(1\minus{}\frac{1}{n^4}\right)^n\cdot \left(1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}$

It is a famous inequality that :

$ \left(1\plus{}\frac{1}{n}\right)^n\le \left(1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}$

So it'll be enough to prove that :

$ \left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}\le \left(1\plus{}\frac{1}{n\minus{}1}\right)^{n}\cdot \left(1\minus{}\frac{1}{n^4}\right)^n$

which i think it is 1990 , Romanian national olympiad


calculus
limit
trigonometry
geometry
inequalities
real analysis
real analysis theorems

Many elementary calculus book prove $ \lim_{x\to0}\frac {\sin x}x \equal{} 1$ by using the area method to get $ \frac12\sin x < \frac12x < \frac12\tan x$. But could you use other ways to prove the inequality, or other methods to prove the limit identity?

It depends on what definition you use. For previous discussions, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=165805]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=120492]here[/url].


I heard this cool riddle today.

There is a king, a queen, 2 sons, 2 daughters, a dog, and a dog keeper on one side of a river.  If the king is with a daughter and the queen isn't there, the king eats the daughter.  If the queen is with a son and the king isn't there, the queen eats the son.  If the dog isn't with the dog keeper the dog eats everyone it is with.  The king, queen and dog keeper can row the boat.  The boat can hold 2 objects.  Get everyone across the river.

[hide]dog and dog keeper go, leave dog
dog keeper takes a son, brings back dog leaving son1
dog keeper and dog get off replaced by son2 and king, son2 left
king and queen return, king gets off
queen returns and gets off, dog and dog keeper go
both get off replaced by king
king picks up queen and then gets off on other side
queen returns to pick up d1 and both get off replaced by dog and dog keeper
dog replaced by d2 who is then left
keeper returns, picks up his dog and all are safely across[/hide]

[quote="LynnelleYe"]The king eats both daughters, or one of them? Same with the queen and the sons?[/quote]

If there are 2, they eat 2; if there is only 1, they eat 1.  Anyways, you don't want anyone to be eaten.

I know...just wanted to be clear on that...

Treething wrote:[hide]
Bring the dogkeeper across the river so the dog eats everybody else. Everybody but the dogkeeper should be in the dog's stomach. Now bring the dogkeeper back to the dog, and bring the dog and the dogkeeper back over.
[/hide]

Oops...everyone should be alive when you get over

hinteaten is dead


geometry
angle bisector
geometry unsolved

Given triangle $ ABC$.From midpoints of $ AB$ and $ AC$ drawn lines perpendicular to bisectors of angles $ C$ and $ B$ respectively.Let these two lines intersects at $ S$.prove that the line joining $ S$ and midpoint of $ BC$ is parallel to bisector of angle $ A$.

$ M,N,L$ are the midpoints of $ AB,AC,CB.$ Let $ M'$ and $ N'$ denote the feet of the angle bisectors of $ \angle NML$ and $ \angle MNL$ in the medial triangle $\triangle  MNL.$ Clearly $ MM'$ and $ NN'$ are parallel to the angle bisectors of $C$ and $ B,$ since $ \triangle LNM$ and $ \triangle ABC$ are centrally similar. Therefore, $ SM \perp MM'$ and $ SN \perp NN'$ $\Longrightarrow$ $S$ is the L-excenter of $ \triangle LNM$ $\Longrightarrow$ $ LS$ bisects $ \angle NLM$ $\Longrightarrow$ $LS$ is parallel  to the angle bisector of $A.$


function
linear algebra
matrix
probability
probability and stats

if we considere a system that has an entry random process (with discret time n )$ E(n)$ that takes the values $ 0$ and $ 1$, and a out random function $ X(n)$ we know that $ P(E(n)\equal{}0)\equal{}P(E(n)\equal{}1)\equal{}\frac{1}{2}$ for all n,and if $ E(n)\equal{}0$ then $ X(n)\equal{}0$ if not $ X(n)$ takes altenativly the values $ 1$ and $ \minus{}1$.
for example if $ E$ takes the values $ 0,1,0,1,0,1,1$ then $ X$ takes the values $ 0,1,0,\minus{}1,0,1,\minus{}1$ or $ 0,\minus{}1,0,1,0,\minus{}1,1$
The question is to find the autocorrelation of $ X$ and its mean.
they gave us this hint :"try to find a Markov chain with 4 states, and find the stationary distribution",my problem is that i have used a Markov chain with 3 states  (.. am not sure of the results ), and i can't see how to use  a Markov chain with 4 states

This is the transition matrix;
$ \left( \begin{array}{cccc}
\frac{1}{2} & 0 & 0 & \frac{1}{2} \\
0 & \frac{1}{2} & \frac{1}{2} & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{2} \\
0 & \frac{1}{2} & \frac{1}{2} & 0  \\
\end{array}\right)$,
space of states is $ \{(E(n),X(n)): n\in\mathbb{N}\}$, the first row is the state $ (0,1)$, second $ (0,\minus{}1)$, third $ (1,1)$, and fourth $ (1,\minus{}1)$ ( same goes for columns ). I'll explain just the probabilitys of the first row;
so in the last outcome, we had $ (0,1)$, that is variable $ E$ was zero and $ X$ was $ 1$, if again $ E$ turns out to be $ 0$ we procede to state $ (0,1)$, with probability $ \frac{1}{2}$, if $ E$ is $ 1$, then because $ X$ was $ 1$, and as you said $ X$ takes vaules $ 1,\minus{}1$ alternativly, we go to state $ (1,\minus{}1)$, again with probability $ \frac{1}{2}$. The rest is more or less the same.
How did you construct the transition matrix?

i have construct this Matrix only from $ X$ because we have to study  the autocorrelation of $ X$....any way it's not true   :oops:  but i still can't see how can you compute the autocorrelation of $ X$

any answer for the autocorrelation ? this actually the part that i hated  :(

I additionally assume that the first non-zero $ X_{t}$ is equally likely to be $ 1$ or $ -1$.

So:
$ E(X_{t})=0$ for all $ t$.
$ Var(X_{t})=E(X_{t}^{2})=E(E_{t}^{2})=E(E_{t})=1/2$.
$ Cov(X_{t},X_{t-k})=E(X_{t}X_{t-k})=P(X_{t}=1\cap X_{t-k}=1)+P(X_{t}=-1\cap X_{t-k}=-1)-P(X_{t}=-1\cap X_{t-k}=1)-P(X_{t}=1\cap X_{t-k}=-1)$.
$ P(X_{t}=1\cap X_{t-k}=1)=P(X_{t}=1\cap X_{t-k}=1|E_{t}=1\cap E_{t-k}=1)\cdot P(E_{t}=1)\cdot P(E_{t-k}=1)=1/4\cdot P(X_{t}=1\cap X_{t-k}=1|E_{t}=1\cap E_{t-k}=1)$.
The first probability ($ P(X_{t}=1\cap X_{t-k}=1|E_{t}=1\cap E_{t-k}=1)$) is the probability that during $ k-1$ trial between time $ t$ and time $ t-k$ there will be an odd number of $ E_{n}=1$. This probability is equal to $ 0.5$ for all $ k>1$ and it is equal to $ 0$ for $ k=1$.
As a final result we get:
$ Cov(X_{t},X_{t-k})=0$ for $ k>1$
$ Cov(X_{t},X_{t-1})=-1/4$ for $ k=1$
And $ Corr(X_{t},X_{t-k)}$ is equal to $ 0$ (for $ k>1$) and $ -1/2$ (for $ k=1$)

but the hint ?! seems like this is a bad problem.. :wink: the question and the hint have not the same goal ...am i right ? or it's just a misunderstanding ?

There are often many ways to deal with the problem. I thought that the direct approach (without Markov) would be easier. I'm sure the author of the question meant another approach.


trigonometry algebra polynomial inequalities

All arguments of sine are in degrees.
(sin1)(sin3)(sin5)...(sin89) = $ \frac{1}{2^n}$

What is n?

[hide="Hint"]
$ \sin\theta\sin\phi \equal{} \dfrac{\cos(\theta\minus{}\phi)\minus{}\cos(\theta\plus{}\phi)}{2}$.[/hide]

I think you would rewrite it as (sin1sin2sin3sin4...sin89)/(sin2sin4sin6...sin88).

the theres alot of double angles if you convert some of the sines to cosines

[hide="Alternately..."] Find a polynomial whose roots are $ \sin 1, \sin 3, \sin 5 ... \sin 89$.  [hide="And note that..."] These values are also $ \cos 1, \cos 3, \cos 5, ... \cos 89$ and that you know what the first and last coefficients of the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomials[/url] are :) [/hide] [/hide]


geometry
blogs
advanced fields
advanced fields theorems

Sorry, but I didn't know where to post this thread. Please feel free to move it around. But do you know any good differential geometry textbook for beginners who don't know anything about it?

Please recommend me some good books!

Thanks

If you are beginner you can see the book Elementary Differential Geometry by Barrett O'Neill this is [url=http://djvu.504.com1.ru:8019/WWW/f3731e7a1fa0ed79bb466b4c36cac9e3.djvu]download link[/url]
There is some others good book about differential geometry as
Differential Geometry of Curves and Surfaces by Manfredo Do Carmo (See my blog)
[url=http://djvu.504.com1.ru:8019/WWW/89189943e5404acb74304f5a4fdd31cc.djvu]Differential Geometry and Its Applications by Oprea[/url]
[url=http://djvu.504.com1.ru:8019/WWW/a8ef32500d5bd4e3a53bb8e09cfa2cef.djvu]Differential Geometry in the Large[/url]
And there is many good book about this subject you can try by google.


[quote="nkht-tk14"]Let $ a_1 , a_2 , a_3, a_4 , a_5$ be nonnegative real numbers
        Prove that

        $ \frac { a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5}{5}$ $ \geq$ $ \sqrt [5]{a_1.a_2.a_3.a_4.a_5}$  + $ \frac { (\sqrt {a_1} \minus{} \sqrt {a_2})^2 \plus{} (\sqrt {a_2} \minus{} \sqrt {a_3})^2 \plus{} (\sqrt {a_3} \minus{} \sqrt {a_4})^2 \plus{} (\sqrt {a_4} \minus{} \sqrt {a_5})^2}{20}$
                Argady , Darij, [i][b]Virgil Nicula [/b][/i] .... can you help me? :ninja:[/quote]
By AM-GM we obtain:
$ \frac { a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5}{5}\minus{}\sqrt [5]{a_1a_2a_3a_4a_5}\equal{}$
$ \equal{}\frac { (\sqrt {a_1} \minus{} \sqrt {a_2})^2 \plus{} (\sqrt {a_2} \minus{} \sqrt {a_3})^2 \plus{} (\sqrt {a_3} \minus{} \sqrt {a_4})^2 \plus{} (\sqrt {a_4} \minus{} \sqrt {a_5})^2\plus{}(a_5\minus{}a_1)^2}{20}\plus{}$
$ \plus{}\frac{a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}a_5\plus{}\sqrt{a_1a_2}\plus{}\sqrt{a_2a_3}\plus{}\sqrt{a_3a_4}\plus{}\sqrt{a_4a_5}\plus{}\sqrt{a_5a_1}\minus{}10\sqrt [5]{a_1a_2a_3a_4a_5}}{10}\geq$
$ \geq\frac { (\sqrt {a_1} \minus{} \sqrt {a_2})^2 \plus{} (\sqrt {a_2} \minus{} \sqrt {a_3})^2 \plus{} (\sqrt {a_3} \minus{} \sqrt {a_4})^2 \plus{} (\sqrt {a_4} \minus{} \sqrt {a_5})^2}{20}.$

[quote="nkht-tk14"]Let $ a_1 , a_2 , a_3, a_4 , a_5$ be nonnegative real numbers
        Prove that

        $ \frac { a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5}{5}$ $ \geq$ $ \sqrt [5]{a_1.a_2.a_3.a_4.a_5}$  + $ \frac { (\sqrt {a_1} \minus{} \sqrt {a_2})^2 \plus{} (\sqrt {a_2} \minus{} \sqrt {a_3})^2 \plus{} (\sqrt {a_3} \minus{} \sqrt {a_4})^2 \plus{} (\sqrt {a_4} \minus{} \sqrt {a_5})^2}{20}$
                Argady , Darij, [i][b]Virgil Nicula [/b][/i] .... can you help me? :ninja:[/quote]
By AM-GM we can prove general problem:
Let $ a_1,a_2,...,a_n>0$. Prove that:
$ \frac{1}{n}\sum_{i\equal{}1}^{n}{a_i}\minus{}\sqrt[n]{\prod_{i\equal{}1}^{n}{a_i}} \ge \frac{1}{n^2} \sum_{i\equal{}1}^{n} (\sqrt{a_i}\minus{}\sqrt{a_{i\plus{}1}})^2, a_{n\plus{}1}\equal{}a_1$

Why we can expand according AM-GM like that?

(a+b+c+d+e)/5-(a*b*c*d*e)^-5=......?

[quote="NguyenDungTN"]
By AM-GM we can prove general problem:
Let $ a_1,a_2,...,a_n > 0$. Prove that:
$ \frac {1}{n}\sum_{i \equal{} 1}^{n}{a_i} \minus{} \sqrt [n]{\prod_{i \equal{} 1}^{n}{a_i}} \ge \frac {1}{n^2} \sum_{i \equal{} 1}^{n} (\sqrt {a_i} \minus{} \sqrt {a_{i \plus{} 1}})^2, a_{n \plus{} 1} \equal{} a_1$[/quote]
I think, $ \frac {1}{n}\sum_{i \equal{} 1}^{n}{a_i} \minus{} \sqrt [n]{\prod_{i \equal{} 1}^{n}{a_i}} \ge \frac {1}{n(n\minus{}1)} \sum_{i \equal{} 1}^{n} (\sqrt {a_i} \minus{} \sqrt {a_{i \plus{} 1}})^2$ better. :wink:

[quote="greentreeroad"]Sorry, I still don't understand why you can do that[/quote]
What do you mean? Maybe are you mean, why

$ \frac{a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}a_5\plus{}\sqrt{a_1a_2}\plus{}\sqrt{a_2a_3}\plus{}\sqrt{a_3a_4}\plus{}\sqrt{a_4a_5}\plus{}\sqrt{a_5a_1}\minus{}10\sqrt [5]{a_1a_2a_3a_4a_5}}{10}\geq0$
is true?
If so, it's true by AM-GM. :wink:

No, in your previous post, I understand the second >= sign by AM-GM as you state
but I don't know about the first equal sign
(a+b+c+d+e)/5-(abcde)^0.2=....?

[quote="greentreeroad"]
but I don't know about the first equal sign
(a+b+c+d+e)/5-(abcde)^0.2=....?[/quote]
We obtain the  equality after expansion !

but still, how do you manage to expand like that

Oh, I got that. But how do you think out that. It seems quite wonderful
Is this way work for n numbers?

[quote="greentreeroad"]Is this way work for n numbers?[/quote]
Yes, you are right! :)

But I think when change to n numbers, the place for 20 should become 4n instead of n(n-1)!
Am I wrong?


Find all real positive $x,y,z$ such that $(x+1)(y+1)(z+1)=2xyz$

I think many answer , like when $x=2$ we have $3(y+1)(z+1)=4yz$ or $y=\frac{3(z+1)}{z-3}$ from which we see that when $z>3$ , there is always a $y$ satisfy it  . Are you sure it is positive real instead of positive INTEGER ?

Well, if it is only integers, I've found only 6 pairs thus far:
$(x,y,z) = (3,4,5)$ in any order

yes it is only positive integers.
Melissa could you please explain how you got your answer.

I'm not even sure if the pairs that I found are the only possible ones, but I simply found them by letting $x+1=y$, $y+1=z$ and $z=2x$.

Use some estimates. Without loss of generality, we can assume that $x \geq y \geq z$. We need \[ \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)\left(1 + \frac{1}{z}\right) = 2.  \] Now, what happens if $z \geq 4$? What are the possible values of $z$? Do you see how to finish it now? :)

[quote="maokid7"]no i still have no idea on how to solve this.[/quote]

from the condition given by Arne , we have

$\frac{1}{x}\leq \frac{1}{y}\leq \frac{1}{z}$

so $2=(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})\leq (1+\frac{1}{z})^3$.......(i)

For any $z\geq 4$ , since $(1+\frac{1}{z})^3\leq \frac{125}{64}<2$ contradict with (i) . This means there is no solution for $z\geq 4$ . Hence we only have to check for the solution $z=1,2,3$ which is just case work . :)


Can any ones here post APMC ( Austria - Poland Maths Contest )
problems in 2001 and solutions ?

hello, a few problems can be found here
http://www.mimuw.edu.pl/~chel/Olimp/olympiads.html
Sonnhard.

oh,thank you so much.
Can you tell me where i can find it 's solution ?


Use Cauchy-Schwartz :

\[ LHS\ge \frac{(a\plus{}b\plus{}c)^2}{4(a\plus{}b\plus{}c)}\equal{}RHS\]

It is too easy for this forum .

Don't post the same problem .

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=300502[/url]

Japanese Communities Modeartor

kunny

[quote="Pirkuliyev Rovsen"]Kolmogorov`s  competition
Prove that if a, b and c are positive numbers,then
a^2/(2a+b+c) + b^2/(a+2b+c) + c^2/(a+b+2c)≥(a+b+c)/4   thanks :wink:[/quote]
Prove that if $a, b $ and $ c $  are positive numbers,then\[\frac{a^2}{2a+b+c} + \frac{b^2}{a+2b+c}+ \frac{c^2}{a+b+2c}\ge \frac{a+b+c}{4 }.\]


Here is sum on clocks:

A man goes out between $ 4$ and $ 5 P.M$. and returns between $ 5$ and $ 6PM$ to find that the hands of the clock had exactly changed places. When does he go out?

Thank you.

[hide] Yeah! How did you do it?
Please explain[/hide]

[hide="solution"]
Let the angle between the hour hand and minute hand originally be B(radians).
Basically,the hour hand travels B degrees and the minute hand travels (2*II - B)
II stands for pi=3.14..

for the hour hand:
in B radians it denotes (B*12*60)/(2*II) minutes     -----1

for minute hand
in (2*II - B) radians it denotes (2*II - B)*60/(2* II) minutes   ---------2

Equating 1 & 2
B=2*II/13
It corresponds to 27.69 degrees   (not needed)
For hour hand it corresponds to 55.384 minutes (not needed)

Originally:
Let A be the angle between the hour hand and 3'O clock (ie the horizantal) originally
Therefore the minute hand is at (A+B+II/2) from the vertical.
it corresponds to 60*(A+B+II/2)/(2*II) minutes   (this denotes minutes past 4'O clock)       -----3

The hour hand denotes 2*II/12 radians for 60 minutes.
The angle it makes with 4'O clock is (A- ( 2*II/12))
It corresponds to (A- (2*II/12)) *60 * 12/(2*II) minutes (this denotes minutes past 4'O clock)             ---------4

Equating 3 & 4
A=69* II/286
This corresponds to 1.447 hours.

Therefore the original time is 1.447 hours past 3'O clock.

Therefore the original time is 4:26.85 pm   (after converting 0.447 into minutes)
[/hide]

[quote]minute hand travels (2*II - B) [/quote]

Why did you write so? Why is $ \pi$ being used here?

The minute hand comes to the place where the hour was originally.
Originally,the minute hand would be betwwen 5 & 6.
The hour hand would be between 4 & 5.
2 * II (radians) denotes the 360 dgrees.

I appreciate your calculations. In the exam, to save time, just try that on your wrist watch (after you have noted the current time in a corner of your paper :rotfl: ). That is why I always recommend wearing an analogue watch rather than a digital one! :blush:

This is a nice way of doing it.
But for the purpose of exams, I would require an easy solution using  algebra, which can be understood by any  other 8th or 9th grader.
Can anyone post such a solution?
Thank you.

Second time:
Hrs hand b.w 5 and 6
mins hand b/w 4 n 5 (since that is whr the hr hand was earlier)
=> that hrs hand will be b/w 5+(4/12) and 5+(5/12)
but this must be the position of minutes hand when the time was first noted
=> minutes hand b/w 5+(4/12) and 5+(5/12)
=> b/w 26.66 and 27.05 minutes
=> b/w 4:26:40 and 4:27:03 (in HH:MM:SS format)

im pretty sure that in most cases.. a more accurate answer is not required
though all u need for a more accurate answer.. is apply the above twice


Let be given a quadrilateral $ ABCD$. Prove that the Euler circles of triangles $ ABC$, $ BCD$, $ CDA$ and $ DAB$ pass through a common point.

Also posted by me at [url]http://www.mathlinks.ro/viewtopic.php?t=150333[/url]. Anyway no solution has been posted yet, so I propose continuing the discussion here, for not bumping the older topic.

Thank you very much dear [b]pohoatza[/b]! I have just found a solution for this problem, here it goes

Denote by $ M$, $ N$, $ P$, $ Q$, $ R$, $ S$ be the midpoints of the segment $ BC$, $ CA$, $ AB$, $ CD$, $ DB$ and $ DA$, respectively. Let $ K$ be the intersection of the Euler circles of triangles $ ABC$ and $ DBC$.

We have
\begin{eqnarray*} \angle (KN, KQ) & = & \angle (KN, KM) + \angle (KM, KQ) \\
& = & \angle (PN, PM) + \angle (RM, RQ) \\
& = & \angle (BC, AC) + \angle (DC, BC) \\
& = & \angle (DC, AC) = \angle (SN, SQ)\end{eqnarray*}
It implies that $ K$ also lies on the Euler circle of triangle $ ADC$. Similarly, it also lies on the Euler circle of triangle $ ADB$. Hence the Euler circles of triangles $ ABC$, $ BCD$, $ CDA$ and $ DAB$ are concurrent as desired.


One hundred people were surveyed. Of these, $ 87$ indicated they liked Mozart and $ 70$ indicated they liked Bach. What is the minimum number of people surveyed who could have said they liked both Mozart and Bach?

[b][size=200][hide = SEE THIS][color=#0ff]OKAY I WILL HIDE SEE I DID JUST NOW[/color][/hide][/size][/b]

[quote=Arthurwei][quote=Residual5000]The minimum number of people who like both Mozart and Bach is achieved when the number of people who like Mozart but not Bach is maximized. There are $ 30$ people who do not like Bach. If all these people like Mozart, then the number of people who like Bach and Mozart is $ 87\minus{}30\equal{}\boxed{57}$.[/quote]

[size=200][color=#0ff]HIDE ANSWERS PLEASE!!!!![/color][/size][/quote]

Do you think they will see that they posted that in 2009

Additionally, the last time they visited AoPS was in 2011.

Wow chill guys they were just trying to help but yeah I don’t think they will see the post

[quote=vectorfunction]Bach is way better than Mozart because:

A) He created more pieces than Mozart

B) He makes better pieces than Mozart

C) His older brother's wedding song was Canon in D by Johanan Pachelbel, and Pachelbel made it as a wedding gift to Bach's older brother.

That is why [color=#0f0]Bach[/color] is better than Mozart[/quote]
Sorry, Bach is well past his expiry date by now. He hasn't been creating more pieces than Mozart since 1749.


conics
ellipse
geometry
geometry proposed

In a given ellipse, major axis $ |A_{1}A_{2}| \equal{} 2a$, focal segment $ |F_{1}F_{2}| \equal{} 2c$ ($ a,c$ are constant). $ P$ is an arbitrary point on the ellipse (but $ P$ is different from $ A_{1}$ and $ A_{2}$). $ D$, $ E$ on $ [F_{1}P],[F_{2}P]$ respectively,  $ DE//BC$ and $ DE$ is tangent of incircle of triangle $ F_{1}F_{2}P$. Prove that $ |DE|$ is constant. (That is, $ |DE|$ is independent from position of $ P$).

Lokman GÖKÇE (Adana)

Name $ r'$ the radius of the excircle of $ \Delta PF_1F_2$ tangent to $ F_1F_2$ and the extensions of $ PF_1$ and $ PF_2$

From triangle similitude we get

$ \frac {DE}{F_1F_2} \equal{} \frac {r}{r}'$ ---> $ DE \equal{} \frac {2c.r}{r'}$ but

$ \frac {r}{r'} \equal{} \frac {s \minus{} F_1F_2}{s}$

$ s$ is the semiperimetr of $ \Delta PF_1F_2$ , $ s$ is constant because $ PF_1 \plus{} PF_2$ is constant, therefore we deduce that the lenght $ DE$ is constant.


Sorry i've asked so many questions already, but the community is so helpful and I believe this is my last question, but I had a problem that I don't quite remember but it asked me to state the domain of it.
The question was a radical over a fraction, but I'm not asking to solve it (since i don't remember it) i'm just hoping someone can tell me what exactly does Stating the domain mean. If someone could give an example of how to solve a question in which it asks you to state the domain that would be extremely helpfull

[url=http://en.wikipedia.org/wiki/Domain_%28mathematics%29]Domain[/url]
This tells you most everything you need to know.

Stating the domain just means find the domain.

[quote="hunter34"]http://en.wikipedia.org/wiki/Domain_(mathematics)

This tells you most everything you need to know.

Stating the domain just means find the domain.[/quote]

That's my problem, I don't know what or how to find the domain, or what it is.
Also, the link didn't work

Fixed the link.

[quote="Wikipedia"]
The domain of a given function is the set of "input" values for which the function is defined.
[/quote]

This input values are of course numbers.

Domain just means all possible $ x$ values

Let's say we had the function $ f(x) \equal{} \frac{1}{x}$, well you can see right away that $ x \ne 0$ otherwise the function is undefined.

So the domain of the function is $ x \in \mathbb{R}, x \ne 0$, or simply $ x \ne 0$

Now let's say we had the function $ f(x) \equal{} \sqrt{x}$. $ x$ must be greater than or equal to $ 0$, so the domain of this function is $ x \ge 0$

Here's one to try

What is the domain of the function $ f(x) \equal{} \frac{\sqrt{x \minus{} 3}}{x \minus{} 7}$


Hi,

I was just wondering the average scores needed to get a decent place in NJ mathcounts. By decent I mean, top four or almsot top four.

Two years ago, I placed 3rd with a 38 pre-countdown.


geometry
circumcircle
parallelogram
rectangle
complex numbers
cyclic quadrilateral

The diagonals $AC$ and $BD$ of a cycilc quadrilateral $ABCD$ intersect at a point $E$. Suppose $O$ is the centre of the circle. If the mid-points of $AD,BC \ and \ OE$ are collinear, prove that $AB = CD$

[color=red][Moderator edit: The problem is wrong. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=38712 .][/color]

I believe your diagram is slightly inaccurate.  But anyways the result holds if either

1) AD || BC

or

2) AC and BD are perpendicular

[quote="probability1.01"]I believe your diagram is slightly inaccurate.[/quote]
You 're absolutely right!

In fact, I didn't use some geometric method to scetch it  ;)

I just found a case for which AB was not equal to CD

[quote="pontios"]:huh:[/quote]
The problem is indeed wrong. I have proved that either AD=BC or the diagonals of the quadrilateral are perpendicular to each other. The picture above illustrates the latter case.
By the way, my proof uses complex numbers, so if anyone finds a purely geometrical solution, please post.

[u]The corrected enunciation.[/u]
Let $ABCD$ be a cyclic convex quadrilateral, the its circumcircle $w=C(O,R)$,
the intersection $E\in AC\cap BD$, the middlepoint $F$ of the segment $[EO]$
and the middlepoints $M,N$ of the sides $[AB],[CD]$.
Prove that $F\in MN\Longleftrightarrow AB\parallel CD\ \vee \ AC\perp BD$.

[u]The my solution.[/u]
Let $U,V$ be the middlepoints of the diagonals $AC,BD$ respectively
and let $G$ be the centroid of the quadrilateral. Thus, $GU=GV,G\in MN$
and the quadrilateral $EUOV$ is cyclically ($OU\perp AC,OV\perp BD$).
[u]The first case:[/u] $G\equiv F$. In this case the quadrilateral $EUOV$ is a cyclical parallelogram,
i.e. $EUOV$ is a rectangle$\Longrightarrow EU\perp EV\Longrightarrow AC\perp BD$.
[u]The second case:[/u] $G\not\equiv F$. In this case $F\in MN\ (G\in MN,FG\perp UV)\Longleftrightarrow$
$MG\perp UV\Longleftrightarrow MU=MV\Longleftrightarrow BC=AD\Longleftrightarrow AB\parallel CD$.

[u]Remark.[/u] Let $ABCD$ be a convex cyclic quadrilateral for which $AD\ne BC$.
Then the points $M,N,F$ are collinearly if only if $AC\perp BD$.


[quote="makar"]There are p identical beads of color red,q identical beads of color blue, r identical beads of color green.
(i)find number of circular arrangements taking all of them at a time.
(ii)find number of necklace which can be formed by taking all of them at a time.[/quote]
It is a long time but please provide the answer to this question.

Wait, aren't part a and part b the same? If not, please state what "necklace" is in this problem.

For part a, let's fix a position $A$ to be red. Then there are $\dfrac{(p+q+r-1)!}{(p-1)!q!r!}$ ways to arrange the rest according to multiple identical items permutation. However, we counted each arrangement $p$ times, once for each red bead (as the one fixed). So we must divide our result by $p$, which gives $\boxed{\dfrac{(p+q+r-1)!}{p!q!r!}}$ ways to arrange the beads in a circle.


Hi guys. I'm a 5th grader who will be 6th grade and a MathCounts

participant in Febuary (or whenever it starts).

I was wondering, for 2008 (or other years), if people could tell

me their state, and what the qualifying score was in that state.

Thanks!!!

For making Nats in a state.

And it varies each year per state. Some states have higher qualifiers, and some don't. Depends on how good the people in your state is.

Hmm... in Massachusetts, the 2008 qualifying score for states was 38. (Countdown is NOT official). But keep in mind that Massachusetts got 4th in the nation at the National competition as well! :)

Delaware probably won't help you, but the 4th placer got a...19. Yes, it's sad, but he's a beast at CD.

I think 3rd and above got like a 28 or above.

Hm...really, just tell us your state because cutoffs range from 0-46... :rotfl:  :rotfl:  :rotfl:

I believe fourth place in Kansas had a score of 35 but since countdown IS official it didn't really matter. I remember one kid who got 7th place written but made it up to second place through countdown.

Alabama:
1st 31
2nd 29
3rd 28 (:D)
4th, 5th, and 6th tied with 27.

CD is not official at any level in our state or else I would have won everything, but alas I won nothing (except for gift cards to the movies).

Alabama= 43rd nats :blush:

I'm from Hawaii and I think the 4th place guy had around a 23.

I think Colorado is sort of average as a state because we got 24th at nats.

I think our 4th place got a 31 or so.

I would not worry about qualifying scores.  The scores needed to qualify for state or nationals change every year depending on the difficulty of the tests.  What would make it one year might not even come close the next year.  A good indication of the difficulty is the school round for the year.  If it is difficult, expect the other tests to be difficult as well.  It could be any part of the test that is difficult.  This year the chapter sprint had 1-3 hard problems, but the target were extremely easy.  The team round was the killer.  At nationals this year there were 2-3 hard or tricky targets but the team round was extremely hard.  Only one team out of 57 got a perfect score.

Instead concentrate on learning as much as you can about as many topics areas as you can.  The trouble with qualifying scores is if you can get that score by the end of the year you think you are in.  But it does not happen that way.  You have to produce on the day of the test.  If you do not get it done, then it is wait until next year.

Concentrate on your weak areas and topics.  When ever you take and old test don't worry about your score.  worry about what you are missing.  Make sure that you go over every problem that you miss.  These should give you an idea of what your weak areas are.  Keep practicing them all year and they won't be weaknesses any more.  Good luck in your first year of competition in Mathcounts.

actually last year and this year, the school rounds were about the same difficulty and this year's may even have been a little more difficult, but last years chapter was difficult relative to other chapters and this years was easy relative to other chapters.


[quote="Ignite168"]SUPERBunny consumes carrots at the rate of 10 carrots per second. SUPERDUPERBunny consumes carrots at the rate of 10 carrots per nanosecond. In a eating contest of a google carrots, how much longer will it take SUPERBunny than SUPERDUPERBunny to finish? Express your answer in scientific notation[/quote]

Just making sure, how long is a nanosecond?

That's part of the problem, google if you must, but try to use parts of the word. Just like that MathCounts problem about abscissa or however you spell it, you can't ask on the test.

[quote="SplashD"]Just making sure, how long is a nanosecond?[/quote]
It's one-billionth of a second. :D

[quote="Ignite168"]SUPERBunny consumes carrots at the rate of 10 carrots per second. SUPERDUPERBunny consumes carrots at the rate of 10 carrots per nanosecond. In a eating contest of a google carrots, how much longer will it take SUPERBunny than SUPERDUPERBunny to finish? Express your answer in scientific notation[/quote]
Well if you're not going to specify what a nanosecond is then you should at least spell googol right...

Google$\neq$Googol.  That is how googol is spelled.  Nanosecond is $10^{-9}$ seconds.


function
complex analysis
real analysis
real analysis unsolved

I got this problem today, and can't seem to solve it:

If P(x)=Q(x) for any real number, show that P(x)=Q(x) for any Complex number.

[quote="ovidius"]Let $u(X)=P(X)-Q(X)$. If $d=deg(u)$, then $u$ has at least $d+1$ roots (actually $u$ has infinitely many roots, since $u(x)=0 , \forall x\in \mathbb{R}$). This implies $u=0 \Longrightarrow P=Q \Longrightarrow P(x)=Q(x) , \forall x\in \mathbb{C}$.[/quote]
Don't you mean: u has d roots?

That proof only works for polynomials.  In fact, it is true also for any entire function because entire functions have isolated zeroes.


find all $(a,b)$ in natural such that $a+b^2|a^2b-1$. :D  :D

I am trying to prove that there are no more solutions except b=1 and a=k keZ thats(a,b)=(k,1) k any integer

Davron ;)

[quote="dna"]find all $(a,b)$ in natural such that $a+b^2|a^2b-1$.[/quote]Let $b=x^2$. The divisibility now becomes $a+x^4|x^2a^2-1 \Rightarrow a+x^4|(xa-1)(xa+1)$. So $a+x^4=xa-1 \Rightarrow a= \frac{x^4+1}{x-1}$ or the other factor, $a+x^4=xa+1 \Rightarrow a= \frac{x^4-1}{x-1} = x^3+x^2+x+1$. This doesn't cover all of the solutions though, such as the set that Davron suggested, but it is clear that there are infinite possibilities.

for any natural number b take the a as a divisor of $b^5+1$ such that is bigger than $b^2$ .
it is clear that there exist this divisor.
any way it is very easy problem.isn`t it? :roll:


Anyone doing research this summer? If so, on what?

I'll be doing some minor research when school starts, my mentor is Mikael Passare.


Hi,

I was wondering if it was possible to change my username to a nickname rather than using the one based on my real name that I have now.

Thanks

Next time stick your wanted username in this thread. Will go much faster than using PMs. ;)

The username I want to change to is "mathmom"

I PM'd Valentin as well.

Thanks

The gods are with the users today :D
Your username has been changed. Have fun!

[quote="shobber"]So I have this question now: Why can't we change the usernames ourselves?  :?[/quote]

Everything would be messed up. Someone you knew yesterday changed identity and you thought he or she quit. It messes up the contests that other people hold too.


Each of the nine dots in this figure is to be colored red, white or blue. No two dots connected by a segment (with no other dots between) may be the same color. How many ways are there to color the dots of this figure?

[asy]draw((-75,0)--(-45,0)--(-60,26)--cycle);
draw((0,0)--(30,0)--(15,26)--cycle);
draw((75,0)--(105,0)--(90,26)--cycle);
draw((-60,26)--(90,26));
draw((-45,0)--(75,0));
dot((-75,0));
dot((-45,0));
dot((-60,26));
dot((15,26));
dot((0,0));
dot((30,0));
dot((90,26));
dot((75,0));
dot((105,0));[/asy]

[hide="Solution"]The first equilateral triangle can be colored in $ 3!\equal{}6$ ways.

Casework for the second equilateral triangle:

(i)[i] The top vertex of the second equilateral triangle is colored the same color as the bottom-[b]right[/b] vertex of the first equilateral triangle[/i]: If this happens, then there are $ 2$ ways to color the remaining two vertices of the second equilateral triangle.

(ii)[i] The top vertex of the second equilateral triangle is colored the same color as the bottom-[b]left[/b] vertex of the first equilateral triangle[/i]: If this happens, then there is only $ 1$ way to color the remaining two vertices of the second equilateral triangle.

Thus, there are 3 ways to color the second equilateral triangle. Similarly, there are also $ 3$ ways to color the thrid equilateral triangle. This gives a total of $ 6\times{3}\times{3}\equal{}\boxed{54}$ ways to color the figure.[/hide]

To Eugenis:
You said the top vertex of triangle 2 has 2 choices. That is correct, but the entire triangle has 3 choices. If color of top vertex 2=color of btm-left vertex 1, there is 1 way to color the rest of the 2nd triangle. If color of top vertex 2=color of btm-right vertex 1, there are 2 ways to color the rest of the 2nd triangle. I hope that explains your question.

[quote=bigmath][hide=Solution]Consider the diagram from left to right. The first dot has 3 choices, the second dot 2, and the third dot is then uniquely determined. Whatever the first dot was colored, we can color any of the next set of 3 dots that color (3 choices). But then the other two dots in that equilateral triangle are fixed. The same is true for the last equilateral triangle. So, there are $3 \times 2 \times 3 \times 3 = \boxed{54}$ ways.

[asy]
draw((-75,0)--(-45,0)--(-60,26)--cycle);
draw((0,0)--(30,0)--(15,26)--cycle);
draw((75,0)--(105,0)--(90,26)--cycle);
draw((-60,26)--(90,26));
draw((-45,0)--(75,0));
dot((-75,0));
dot((-45,0));
dot((-60,26));
dot((15,26));
dot((0,0));
dot((30,0));
dot((90,26));
dot((75,0));
dot((105,0));
[/asy][/hide][/quote]

I am confused about how you  "color any of the next $3$ dots that color.? By this, are you referring to the second equilateral triangle? If so, choosing any color will not uniquely determine a coloring because two vertices of the triangle may be the same color. For example, if I choose the bottom-right vertex of the second triangle and select the same color as for the top vertex of the first triangle, then there are $2$ choices for the top vertex of the second triangle. Casework bashing yields that there are $3$ ways to color the second triangle, but just coloring a point of the second triangle and stating that the second and third points are uniquely determined is not correct. If you can elaborate on your sol, please post below in the thread.

Do we have to use casework to determine there are 3 ways for the 2nd triangle or is there another way?

How I did it:

Casework on the top:

Notice there are two choices for the top, all different, or middle one different, outer two same, then do casework on the middle and get 24. Same thing for all different, and you will get 30
30+24 = 54


I changed the wording.

If $a * b = \frac{a-b}{a+b}$ and $8 * (2 * x) = \frac{4}{3}$, evaluate $x$.