Which one is larger: $\log_{7}8$ or $\log_{8}9$.

[quote="computer_nuke"][quote="Kurt Gödel"][hide="Nice question"]Let $a=\log 8-\log 7$ and $b=\log 9-\log 8$. Clearly, $a-b = 2\log 8-\left(\log 7+\log 9\right) = \log 64-\log 63 = \log \frac{64}{63}> 0$, so $a > b$. Since $a,b > 0$, we know that $ab > 0$. And because $\log 8 > 0$, it follows that $(a-b)\log 8 > 0$. Summing those two inequalities we get that $(a-b)\log 8+ab > 0$. Therefore, we have that \[\left(\log 7\right)\left(\log 9\right) = \left(\log 8-a\right)\left(\log 8+b\right) = \left(\log 8\right)^{2}-(a-b)\log 8-ab < \left(\log 8\right)^{2}.\] But this means $\log_{8}9 = \frac{\log 9}{\log 8}< \frac{\log 8}{\log 7}= \log_{7}8$.[/hide][/quote]

Nice solution. How did you come up with it? I mean the creation of the numbers a and b and stuff.[/quote]You immediately find the equivalent inequality $\left(\log 7\right)\left(\log 9\right) < \left(\log 8\right)^{2}$. You just know that it's true, because you know that $\log 7 < \log 8$ and $\log 8 < \log 9$, but you feel that the first inequality isn't as "strong" as the latter. If you could prove this, you're done. So you want to say in mathematical language that $\log 8-\log 7 > \log 9-\log 8$...  :)
Anyway, thanks for the compliment!

What about $\log_{x}{x+1}$ or $log_{y}{y+1}$, when $x>y$? (I mean, generalization)

[quote="jhcreinhardt"]$\log8*\log8>\log7*\log9$[/quote]
How did you know this was true? Common sense?

[quote="jhcreinhardt"]is it ok to assume am-gm applies to log?[/quote]

It would be much easier to do if you could assume that.

[quote="jhcreinhardt"]is it ok to assume am-gm applies to log?[/quote]

in this case the logs are positive, so I guess here it's okay.


algebra
polynomial
search
algebra proposed

Let $ \left(F_1,F_2,F_3,...\right)$ be the Fibonacci sequence, defined by the starting values $ F_1 \equal{} 1$ and $ F_2 \equal{} 1$ and the recurrence equation $ F_{n \plus{} 2} \equal{} F_{n \plus{} 1} \plus{} F_n$ for all $ n\geq 1$.
Let $ P\left(x\right)$ be a polynomial of degree $ 990$ such that $ P\left(k\right) \equal{} F_k$ for all $ k \in\left\{ 992, 993, ... , 1981, 1982\right\}$.
Show that $ P\left(1983\right) \equal{} F_{1983} \minus{} 1$.

Using finite differences and the facts that the nth finite difference of an n-degree polynomial is constant and that $f_3 - f_2 = 1$, the result follows immediately.

Think about the generalized finite differences of $P(x)$

You may search in Internet: [url]http://planetmath.org/encyclopedia/LagrangeInterpolationFormula.html[/url]

[quote="JBL"]Using finite differences and the facts that the nth finite difference of an n-degree polynomial is constant and that $f_{3}-f_{2}= 1$, the result follows immediately.[/quote]
can u explain to me? thx!


function
topology
advanced fields
advanced fields unsolved

Let $f: \mathbb R^{n}\longrightarrow\mathbb R^{m}$ be a differentiable function. Let $a,b\in\mathbb R^{n}$. Suppose that $\{Df_{x}|x\in[a,b]\}$ is convex subset of $L(\mathbb R^{n},\mathbb R^{m})$. Prove that there is $x\in[a,b]$ that \[f(a)-f(b)=Df_{x}(a-b)\]

What do you mean by $[a,b]$? Intervals don't really make sense in $\mathbb{R}^{n}$.

[quote="Omid Hatami"]
Suppose that $\{Df_{x}|x\in[a,b]\}$ is convex subset of $L(\mathbb R^{n},\mathbb R^{m})$.

...

Nobody Interested?
[/quote]

It's very strange condition. Can you give an example of function $f$ such that $\{Df_{x}|x\in[a,b]\}$ is a convex set, but isn't a segment?

I think we can prove with Sard's Theorem, that $Df_{x}[a,b]$ is always a 1-manifold, but problem has easier solution.


modular arithmetic
induction
number theory proposed
number theory

Let (a,b)=1 and T is minimal period b mod a ($ a|b^T\minus{}1, a|b^k\minus{}1\Longrightarrow T|k$). Then (T,a)=1 and define k, suth that $ kT\equal{}1\mod a$. It give for n=kT: $ a|b^n\minus{}n$.
It may be when $ (a,b)\not \equal{}1$. Let (a,b)=c and $ a\equal{}ca_1,b\equal{}cb_1, (a_1,b_1)\equal{}1$. Then $ a|b^n\minus{}n$ only if $ c|n$.
If $ (a_1,b)\equal{}1$ and $ T$ is period $ b$ mod $ a_1$, $ kcT\equal{}1\mod a_1$ then for $ n\equal{}kcT$ we get $ a|b^n\minus{}n$.
By concider $ (a_1,c)\equal{}d,a_1\equal{}da_2$, we get, that exist n, suth that $ a|b^n\minus{}n$ for any a,b.

[quote="Rust"]Let (a,b)=1 and T is minimal period b mod a ($ a|b^T \minus{} 1, a|b^k \minus{} 1\Longrightarrow T|k$). Then (T,a)=1 [/quote]
It is not true.
For example $ a\equal{}p^2$ and $ b$ is a primitive root mod a.
Then $ T\equal{}p(p\minus{}1)$
This problem can be proof by Chinese remainder .
Consider the $ n\equal{}p^a$

Yes. If $ b=p_1^{k_1}..p_m^{k_m}b_1,a=p_1^l_1..p_m^{l_m}a_1, (a_1,b_1)=1= (p_i,a_1)=(p_i,b_1)$, then from $ a|b^n-n$
we get $ n=n{}_0n{}_1, n_0=p_1^{l_1}...p_m^{l_m}$ and $ a|b^n-n$ if and only if $ a_1|b_2^{n_1}-n{}_0n{}_1$ were $ b_2=b^{n_0}$.
 Let $ a_2=rad(a_1)$. Then exist minimal period $ T$, suth that $ a_2|b_2^T-1$. Let $ k$ defined by $ kn_0T=1\mod a_2$ ,  ($ (T,a_2)=1$).
It give solution $ a_2|b_2^{n_1}-n{}_0n{}_1$ ($ n_1=kT$). But case, when $ \frac{a_1}{a_2}$ had many prime divisors is difficult. Because chinese theorem does not work for full periods $ b^n-n$ and $ a=p_1^2p_2^2$ may be $ (T_1,T_2)\not =1$.

I thinks i have a solution! we will show that "there is infty n such that $ a|b^n-n,\forall a>1$"
+) a=2, it's trivial
+) Suppose it's true for a. we have $ \exists k,a_1: a|b^ka_1,gcd(a_1,b)=1$. Because $ \varphi(a_1)<a_1\leq a$ so there is infty n>k such that $ \varphi(a_1)|b^n-n$
$ \Rightarrow a_1|b^{b^n-n}-1$
$ \Rightarrow b^na_1|b^{b^n}-b^n$
$ \Rightarrow a|b^{b^n}-b^n$
with infty n. The proof is complete! Sorry if the text have some error :oops:

Let $ a_2 \equal{} rad(a_1)$.

what is rad(a1)?

To type up langtuthanhdong's proof in a polished form:

Let us proceed by induction. The base case $ a\equal{}2$ is trivial. Now, suppose that for all $ k<a$, there exists $ n$ such that $ b^n \equiv n \pmod{a}$. Let $ b\equal{}gb', a\equal{}ga'$, where $ (b',a')\equal{}1$. Then, $ \varphi(a')<a'<a$, so there exists $ n$ such that
\[ \varphi(a')| b^n\minus{}n
\]
So
\[ a' | b^{b^n\minus{}n}\minus{}1
\]
So
\[ b^na' | b^{b^n}\minus{}b^n
\]
But $ a | b^na'$, so
\[ a | b^{b^n}\minus{}b^n
\]
And the induction is complete.

[quote="Phelpedo"]
So
\[ a' | b^{b^n \minus{} n} \minus{} 1
\]
[/quote]

What if $ a\equal{}8$, $ b\equal{}4$? The proof seems to fail for $ (a',g)\neq1$

[quote="N.T.TUAN"]Determine whether or not there exist positive integers $ a$ and $ b$ such that $ a$ does not divide $ b^n \minus{} n$ for all positive integers $ n$.[/quote]

Well it just says to determine, so couldnt we just say "let a=5, b=3, and n=2 QED" and hope for at least partial credit?

We can say more:

[i]"Let $ a,c,d$ positive integer and $ b \in \{\minus{}1,1\}$ fixed, then there exist infinitely many positive integer $ x$ such that $ d \mid a^x\plus{}bx\plus{}c$".[/i]

Have a look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=354533#354533]here[/url]  :wink:

I didn't see a confirmed solution here, so maybe the following will work:

Answer is no, so we can always find such n with a to divide $ b^n\minus{}n$.

We can first consider the case of modulo p.  We have that $ b^n$ has cycle length at most p-1 (and which divides p-1); on the other hand, $ n$ just has cycle length p, hitting all the values.  Thus, we can say that there are $ p\minus{}1$ values $ j$ in modulo $ p(p\minus{}1)$ such that $ b^j \equiv j \pmod{p(p\minus{}1)}$, all of these values $ j$ being different modulo p-1 (so there is a valid j for every modulo p-1 value).

Now I claim that for any $ m>1$, we have p-1 values $ x$ in modulo $ p^{m\minus{}1}(p\minus{}1)$ such that $ b^x \equiv x \pmod{p^{m}}$, and such that the p-1 values are distinct mod p-1.  To do this, we show that for any value $ x$ such that $ b^x \equiv x\pmod{p^{m\minus{}1}}$, we can find $ y\equiv x\pmod{p\minus{}1}$ with $ b^y\equiv y \pmod{p^{m}}$.

We take $ y \equal{} s(p^{m}\minus{}p^{m\minus{}1}) \plus{} x$ (for some $ s$) and so $ b^{y} \equal{} b^{s(p^{m}\minus{}p^{m\minus{}1})\plus{}x} \equiv b^x \equiv x \plus{} s(p^{m}\minus{}p^{m\minus{}1}) \equiv y\pmod{p^m}$ for appropriately chosen $ s$.  Thus, our claim is proven.

Now, we can construct a number $ n$ with $ a|b^n\minus{}n$.
Let $ a\equal{}p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ such that they are ordered $ p_1<p_2 < \cdots <p_k$.
Clearly we can construct $ n$ so that $ p_1^{e_1} | b^n \minus{} n$.
Now, suppose we can construct one for some number of primes $ p_1^{e_1}$ through $ p_i^{e_i}$ dividing $ b^n\minus{}n$.  This will hold for $ n$ restricted within a congruence modulo $ \phi (p_1^{e_1} \cdots p_i^{e_i})$.
For $ p_{i\plus{}1}^{e_{i\plus{}1}}$ to divide $ b^n\minus{}n$, we will need to have $ n \equiv h\pmod{p_{i\plus{}1}^{e_{i\plus{}1}\minus{}1}(p\minus{}1)}$ for some $ h$.  However, we can choose the value of $ h$ in modulo $ p_{i\plus{}1}\minus{}1$ by the claim from before, and we have that $ p_{i\plus{}1}$ is larger than all the prime factors of $ \phi (p_1^{e_1} \cdots p_i^{e_i})$.  Thus, imposing this congruence of $ n\equiv h\pmod{p_{i\plus{}1}^{e_{i\plus{}1}\minus{}1}(p\minus{}1)}$ will have solution.

Thus, we can construct n, so we are done.

Suppose, by way of contradiction, that there exist positive integers $a, b$ such that $a \nmid b^n - n$ for all $n \in \mathbb{N}.$ Then for a fixed $b$, let us examine the [i]smallest [/i]positive integer $a$ such that $a \nmid b^n - n$ for all $n \in \mathbb{N}.$ Notice that trivially $a \ne 1.$ Now, consider the least exponent $d \in \mathbb{N}$ such that $b^{i + d} \equiv b^i \pmod{a}$ for all $i \in \mathbb{N}$, i.e. $d$ is the period of the sequence $b, b^2, b^3, \cdots \pmod{a}.$ Note that if $\gcd(a, b) = 1$, then all terms of this sequence are relatively prime to $a \implies d \le \phi(a) < a.$ Otherwise, if $a$ and $b$ share a common factor, then no term of this sequence is equal to $1.$ Therefore, there are at most $a - 1$ possible values for terms of this sequence, and it follows that $d < a.$

Now, since $d < a$, by the minimality of $a$, there must exist some $k \in \mathbb{N}$ such that $d \mid b^k - k.$ Hence, we can write $b^k = k + cd$ for some $c\in \mathbb{N}_0.$ Then let us take $n = b^k.$ By the definition of $d$, it follows that \[b^n - n = b^{b^k} - b^k = b^{k + cd } - b^k \equiv b^k - b^k \equiv 0 \pmod{a}.\] Hence, $a \mid b^n - n$, a contradiction. $\square$

Cases $a=1,2$ are trivial. Let $a$ be the least such positive integer. Then any prime dividing $a$ also satisfies this so we get a contradiction to the minimality of $a$ unless $a$ is a prime.
Now let $p$ be a prime such that $p$ divides $a$. Put $n$ = $((1-b)+kp)(p-1)+1$. We get that $p$ divides $b^n -n$. Done.(as indeed by minimality $a$ =$p$)
Its wrong dont consider this.

[quote=anantmudgal09]Cases $a=1,2$ are trivial. Let $a$ be the least such positive integer. [color=#f00]Then any prime dividing $a$ also satisfies this[/color] so we get a contradiction to the minimality of $a$ unless $a$ is a prime.
Now let $p$ be a prime such that $p$ divides $a$. Put $n$ = $((1-b)+kp)(p-1)+1$. We get that $p$ divides $b^n -n$. Done.(as indeed by minimality $a$ =$p$)[/quote]
This does not seem true to me! :maybe:

[quote=Dukejukem]Now, consider the least exponent $d \in \mathbb{N}$ such that $b^{i + d} \equiv b^i \pmod{a}$ [color=#f00]for all $i \in \mathbb{N}$. [/color] Because $b\left(b^{\phi(a)} - 1\right)$ is divisible by $a$, we have that \[b^i\left(b^{\phi(a)} - 1\right) \equiv 0 \pmod{a} \implies b^{i + \phi(a)} \equiv b^i \pmod{a}.\][/quote]

If $p$ is a prime dividing both $a$ and $b$, and $v_p(b) < v_p(a)$, then this is not true for $i = 1$.

Oh yes i made some mistake :oops_sign: (must be my lack of sleep) i will edit the previous post and will try it over again. Thanks a lot.

[quote=saturzo]
[quote=Dukejukem]Now, consider the least exponent $d \in \mathbb{N}$ such that $b^{i + d} \equiv b^i \pmod{a}$ [color=#f00]for all $i \in \mathbb{N}$. [/color] Because $b\left(b^{\phi(a)} - 1\right)$ is divisible by $a$, we have that \[b^i\left(b^{\phi(a)} - 1\right) \equiv 0 \pmod{a} \implies b^{i + \phi(a)} \equiv b^i \pmod{a}.\][/quote]

If $p$ is a prime dividing both $a$ and $b$, and $v_p(b) < v_p(a)$, then this is not true for $i = 1$.[/quote]

Ah, thanks for the catch. I'll edit the proof accordingly.

I don't see this here?

By a trivial extension of [url=http://artofproblemsolving.com/community/c6h16294p113561] 1991 USAMO #3[/url], the sequence $b, b^b, b^{b^b}, \cdots$, is eventually constant modulo $a$. Then, just take a sufficiently large power tower in $b$ for $n$, and it works trivially.

[quote=Dukejukem]Now, consider the least exponent $d \in \mathbb{N}$ such that $b^{i + d} \equiv b^i \pmod{a}$ for all $i \in \mathbb{N}$, i.e. $d$ is the period of the sequence $b, b^2, b^3, \cdots \pmod{a}.$[/quote]

I would like to see the proof of this periodicity when $(a,b)\neq 1$. Certainly, there exist $i,d\in Z_+$ as marked above. And from $b^d$ sequence will be periodic. But will he be periodic before $b^d$? If so, then how do you prove it?

Induction :P. Here's my solution (Hopefully correct): We claim that for each $(a,b),$ there exists some $n$ such that $a \mid b^n-n$. Consider the infinite sequence of positive integers defined as $b_1=b$ and $b_{k+1}=b^{b_k}$. Then the following Lemma is true:-

[b]LEMMA[/b] (USAMO 1991 P3) The sequence $<b_k>_{k \geq 1}$ eventually becomes constant modulo $a$.

[i][u]Proof of Lemma[/u][/i] We use strong induction on $a$. For $a=1,2$ the result is obvious. Suppose our claim is true for $1,2, \dots ,a-1$. Consider the case when $\gcd(a,b)=d>1$. Choose some prime divisor $p$ of $d$. Note that, for sufficiently large $j$, $p^{\nu_p(a)} \mid b_j$. So we can effectively ignore $d$, and assume that $\gcd(a,b)=1$. Then it is well known that $$b_{k+1}=b^{b_k} \equiv b^{b_k \pmod{\phi(a)}} \pmod{a}$$
But, by the induction hypothesis, for sufficiently large $k$, the sequence $<b_k>_{k \geq 1}$ eventually becomes constant modulo $\phi(a)$, giving the desired result. $\Box$

Return to the problem at hand. Then choosing $n=b_k$, for sufficiently large $k$ concludes our solution. Hence, done. $\blacksquare$

[b]REMARK:[/b] Without knowing the Lemma, the problem can be difficult. Unfortunately, this happened with me, as I wasn't aware of it. Here's how I thought of it - It's quite intuitive to think that the answer is not affirmatory. The main difficulty lies in finding the desired $n$. On looking at the problem, one can see that $b^n$ and $n$ are not closely related, unless $n$ is a power of $b$. Then, trying some small cases, and to make things even more symmetric, I thought of introducing towers of $b$. Apparently they worked :D.

We claim that given any positive integers $a$ and $b$, there always exists a positive integer $n$ such that $a\mid b^n-n$.

Note that replacing $a$ with any multiple of $a$ makes the problem strictly harder, so WLOG, we may assume that the prime factorization of $a$ is of the form $p_1^r\cdots p_k^r$ where $2=p_1<p_2<\cdots<p_k$ are all of the primes from $2$ to $p_k$. This won't actually affect the content of the proof, but will make the argument slightly easier to make.

Let $M_i$ be $p_1\cdots p_i$. We claim that for each $1\le i\le k$, there exists integers $n_i$ and $N_i$, the latter of which is a multiple of $N_i$ with no new prime factors, such that $n\equiv n_i\pmod{N_i}$ implies $M_i\mid b^n-n$. We proceed by induction on $i$. The base case of $i=1$ is easy, since we may take $n_1=b$ and $N_1=p_1=2$. Now suppose that the statement is true for some $i$, and we want to show it for $i+1$.

Note that $p_{i+1}-1$ has all prime factors from $p_1,\ldots,p_i$. Thus, by CRT, there is a solution $n_{i+1}$ to the system of congruences
\begin{align*}
n_{i+1} &\equiv n_i\pmod{N_i} \\
n_{i+1} &\equiv n_i\pmod{p_{i+1}-1} \\
n_{i+1} &\equiv b^{n_i}\pmod{p_{i+1}},
\end{align*}
since $p_{i+1}$ is relatively prime to both $p_{i+1}-1$ and $N_i$. Let $N_{i+1}=\mathrm{lcm}(M_i,p_{i+1}-1,p_{i+1})$ and as said above, $n_{i+1}$ is a solution to the given system of congruences. If $n\equiv n_{i+1}\pmod{N_{i+1}}$, then $M_i\mid b^n-n$ by the first congruence and the induction hypothesis. Furthermore, by the last two, we see that $b^n\equiv b^{n_i}\pmod{p_{i+1}}$ and $n\equiv b^{n_i}\pmod{p_{i+1}}$, so $p_{i+1}\mid b^n-n$, as desired.

Thus, by induction, there is some integer $L_1$ (which is just $N_k$) with prime factors only from $p_1,\ldots,p_k$ and some other integer $\ell_1$ (which is just $n_k$) such that
\[n\equiv \ell_1\pmod{L_1}\implies p_1\cdots p_k\mid b^n-n.\]
We claim that for all $1\le j\le r$, there exist integers $L_j$ (with all prime factors from $p_1,\ldots,p_k$, each at least $j$ times, and also divisible by $p_i-1$ for all $1\le i\le k$) and $\ell_j$ such that
\[n\equiv\ell_j\pmod{L_j}\implies p_1^j\cdots p_k^j\mid b^n-n.\]
The base case was established above, so like before, we suppose the result is true for $j$, and we're trying to show it for $j+1$. Let $\ell_{j+1}$ be the solution to all the systems of congruences
\begin{align*}
\ell_{j+1} &\equiv \ell_j\pmod{L_j} \\
\ell_{j+1} &\equiv \ell_j\pmod{p_i^j(p_i-1)} \\
\ell_{j+1} &\equiv b^{\ell_j}\pmod{p_i^{j+1}},
\end{align*}
where $i$ ranges over all integers from $1$ to $k$ inclusive. Note that this only works because the third congruence and the first two conflict over a modulus that is a factor of $L_j$ since $L_j$ has at least $j$ factors of $p_i$ and is divisible by $p_i-1$, and $b^{\ell_j}\equiv\ell_j\pmod{L_j}$. Letting $L_{j+1}$ be the lcm of $L_j$ and all the $p_i^{j+1}$s (the $p_i-1$ if a factor of $L_j$ already), we see that $n\equiv\ell_{j+1}\pmod{L_j}$ means that
\[b^n-n\equiv b^{\ell_j}-b^{\ell_j}\pmod{p_i^{j+1}},\]
so $p_1^{j+1}\cdots p_k^{j+1}\mid b^n-n$, thus completing the induction.

Therefore, there are integers $y$ (which is $\ell_r$) and $Y$ (which is $L_r$) such that
\[n\equiv y\pmod{y}\implies a\mid b^n-n,\]
so we're done.

There is always an integer $n$ that makes $b^n  - n \equiv 0 \pmod{a}$ for given $a$ and $b$.

Let us induct on $a$. For small $a$, such as 1, and 2, the claim holds true.
For $a \geq 3$, assume that for some $b$ for no such $n$ $b^{n} - n \equiv 0 \pmod{a}$. Then, for no $n= b^k$, $b^{b^k} \equiv b^k  \pmod{a}$. But this cannot be true by induction, as for some $k$, $b^{k} \equiv k \pmod{\phi(b)}$.

The answer is no; we claim that for all positive integers $a,b,$ there is some $n$ such that $a\mid b^n-n.$ To prove this, we will fix $b$ and induct on $a,$ the base case being clear.\\\\ Suppose the claim is true for $a=2,3,\dots,k.$\\\\ If $\gcd(b,k+1)=1,$ then let $n=b^m,$ where $$b^{m}\equiv m\pmod{\varphi(k+1)}.$$ (The inductive hypothesis guarantees that a solution exists) \\By Euler's Theorem, $b^n\equiv n\pmod{k+1}.$ \\\\ If $\gcd(b,k+1)>1,$ then let $$d=\prod_{\text{p is prime, }p\mid b}p^{\nu_{p}(k+1)}.$$ Now let $n$ be a solution to the congruence $$b^{n}\equiv n\pmod{(k+1)/d}.$$ Since $\gcd(d,(k+1)/d)=1,$ we have $b^n\equiv n\pmod{k+1},$ as desired.

The answer is no.
Since $\varphi(n)<n$ for all $n>1$, consider $k \geq 0$ such that $\varphi^k(a)=1$, which must exist. Now let $n={^{k}b}$ (denoting tetration), so $b^n={^{k+1}b}$. Then by Euler Totient
$${^{k+1}b}\equiv {^{k}b} \pmod{a} \impliedby {^{k}b}\equiv {^{k-1}b} \pmod{\varphi(a)} \impliedby \cdots \impliedby {^{1}b}=b\equiv 1={^{0}b} \pmod{\varphi^k(a)},$$
which is true as clearly $1 \mid b-1$.

The answer is no.

We claim that $\underbrace{b^{b^{\cdots^b}}}_{n}$ is eventually constant $\pmod a$. We will prove this by induction on $a$. Indeed, the basis of $a=1$ is trivial. Now, assume the result holds for every positive integer less than $a$. We let $f(n)=\underbrace{b^{b^{\cdots^b}}}_{n}$. Let the period of $b,b^2,b^3, \ldots$ mod $a$ be $d$ which is strictly less than $a$. By our hypothesis, $f(k)$ is eventually constant mod $d$ and thus, $b^{f(m)}=f(m+1)$ is eventually constant mod $a$. So, letting $n=f(N)$ for sufficiently large $N$ yields $b^{f(N)}-f(N) = f(N+1)-f(N)$ which is divisible by $a$.

This is quite tricky for TST4; the obvious solution of ``solve the problem for $a = p^k$ and then generalize" doesn't work very well, and you need to tailor it.

The answer is no; fix $a = p_1^{k_1} p_2^{k_2} \cdots p_\ell^{k_\ell}$. We prove that no such $b$ exists for $a =(p_1p_2 \cdots p_\ell)^k$ for all positive integers $k$ by induction, which will imply the result.

Now instead of looking at the primes separately, we look at them [i]together[/i]! The base case $k=0$ is vacuous. For the inductive step, suppose we have an $n_0$ such that $$b^{n_0} - n_0 = c(p_1p_2 \cdots p_\ell)^k$$ for some constant $c$. Then, observe that $$b^{n_0+r\prod_{i=1}^\ell p_i^k(p_i-1)} - n_0 - ip_1^k p_2^k \cdots p_\ell^k(p_1-1)(p_2-1) \cdots (p_\ell - 1) \equiv p_i^k\left(\prod_{j \neq i} p_i^k \right)(c-r(p_1-1)(p_2-1)\cdots (p_\ell - 1)) \pmod {p_i^{k+1}}$$ because the exponent term is congruent to $b^{n_0}$ by Euler totient. On the other hand, by picking a suitable $r$ modulo $p_i$, we can guarantee that $$p_i \mid c - r(p_1-1)(p_2-1) \cdots (p_\ell - 1)$$ and thus $p_i^{k+1}$ divides the original expression. We may collate this across all $p_i$ by Chinese Remainder Theorem to complete the inductive step.

The answer is no. We proceed by induction, suppose that for all $k<a$ it exists an $n$ such that $a\mid b^n-n$, where are going to prove the statement for $a$, Let $\gcd(a,b)=k$, as $\phi(\frac{a}{k})<a$ then
\begin{align*}
\phi(\frac{a}{k}) &\mid b^n-n \\
\frac{a}{k} &\mid b^{\left(b^n-n\right)}-1 \\
\frac{a}{k}\cdot b^n &\mid b^{b^n}-b^n\\
a &\mid b^{b^n}-b^n
\end{align*}
We are done

Note that the problem it´s trivial after seeing [url=https://artofproblemsolving.com/community/c6h16294p113561] 1991 USAMO #3[/url]  :-D


algebra
polynomial
superior algebra
superior algebra unsolved

Why is $ x$ irreducible in $ \mathbb{C}[x,y,z,t]/<xz\minus{}yt>$??

The ideal is homogeneous, so quotienting by the ideal preserves the grading.  There's no way for $ x$ to be a product of two nonconstant polynomials.


$ a,b,c > 0$ and $ k \equal{} \sqrt[3]{abc}$
Prove:
$ \frac{a^2}{k^2\plus{}bc} \plus{}\frac{b^2}{k^2\plus{}ac} \plus{}\frac{c^2}{k^2\plus{}ab} \geq \frac{\sqrt{bc}}{\sqrt{ka}\plus{}a} \plus{}\frac{\sqrt{ac}}{\sqrt{kb}\plus{}b}\plus{}\frac{\sqrt{ab}}{\sqrt{kc}\plus{}c}\geq \frac{3}{2}$

MHS-RHS:

It's homogeneous, so let $ abc \equal{} 1$
It becomes:
$ \sum \frac {1}{a\sqrt {a} \plus{} a} \geq \frac {3}{2}$
But $ \sum \frac {1}{a\sqrt {a} \plus{} a} \geq \frac {(\frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}})^2}{3 \plus{} \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}} \equal{} \frac {1/a \plus{} 1/b \plus{} 1/c \plus{} 2(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})}{3 \plus{} \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}} \geq \frac {3}{2}$
The last ineq is equivalent to:
$ 2(1/a \plus{} 1/b \plus{} 1/c) \plus{} (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \geq 9$ which is just AM-GM


For those of you who aren't familiar with big brother, let me explain.
Its a show on CBS similar to survivor.
13 people move in to a house with no contact to the outside world.
Every week there is a competition for a spot of power called Head of Household.
He/She can then nominate two other players for eviction.
Then, there is another competition between the houseguests for a power called the Power of Veto
The winner can save the nominees from nomination.
Then the other players vote for which nominee they want to eliminate.
go to cbs.com/bigbrother if you want to watch the actual show

Post message if you wanna signup

Player list:
1. Mewto55555
2. Meewhee009
3. zapi2007
4. isabella2296
5. spaceguy524
6. eggylv.999

7 more spaces, yo

I'll play.
Don't be afraid of prods...or modkills.

They will speed up the game and make sure people are active.

I'll join, even though I don't watch it.

I just dont want this topic to sink before it starts

It won't. Also, can you discuss with other people in the game outside of thread?

How do you decide between final two?

sure man but try to keep it in this thread as much as you can to keep it entertaining.

for those signed up, feel free to stragetize now.

By the way, did you watch this year's and how they split up the people by cliques?  I thought that was interesting.

Extra rules with cliques including that HOH's clique can't be nominated and they had the have's and have-not's rule but that's unfeasible for this game.

I join The message is too small. Please make the message longer before submitting.

Blah...
supernerd123 and izzy and zapi....

Please vote for me! I know that izzy will vote for mewto to win, but.....

You never know...

Hey nacho, can you make some questions for our current candidates to answer?

or maybe some questions for us to answer?

or both?

We are still waiting on Supernerd123 and Zapi2007
 :roll:

so we're not going to hear any much?

then I'll ask a question.

the nominees

what was your strategy coming into this game?

blah survive b/c mewto nominated me first found for no reason...

Win every challenge so as to ensure I didn't get out that round; unfortunately winning too many HoH earned me too many enemies.

I would like to announce the winner of Big Brother AoPS right now! For those who are reading this: do not tell other players what happened! Let them read it themselves!
This is brought to you via copy-paste.

supernerd123 voted:
[hide]he had not decided yet[/hide]

zapi2007 voted:
[hide]ill vote for Mewto55555[/hide]

echinodermata voted:
[hide]Ill vote for Eggylv.999[/hide]

meewhee009 voted:
[hide]Mewto55555[/hide]

isabella2296 voted:
[hide]Mewto55555. Congratulations Mewto55555, you are the winner of Big Brother!!!!!!!!!! It was super fun getting to manage this. This thread is not over however, as in a few days i will summarize this entire season from my point of view, which will be interesting as i was able to see all that was going on.[/hide]

Congratulations to the winner of Big Brother! It has been fun, but dont tune out yet, i will be back with something else.

had spaceguy not voted me out, I think I would have at least had last three.
I had a running alliance with all three of the finalists way before everyone started losing, and I believe I would have had a shot at the last HoH competition, seeing that I got there way before mewto did.

What happened to that little advantage the second person (echinodermata) got after the first part of the hoh competition?

When I cast my vote, I thought mewto was going to lose for sure.  He made too many enemies too quickly.

[quote="zapi2007"]

When I cast my vote, I thought mewto was going to lose for sure.  He made too many enemies too quickly.[/quote]

Hehe, I had alliances with zapi, izzy, and meewhee. I figured it would work so long as I tanked the HoH competition in a round when I would have had to nominate one of you; as it turned out it did. Good game eggy.

OVERVIEW OF THE GAME
Okay, when I first started this I didnt expect it to be as effort consuming as i thought, but i constantly had to think of new ideas and scenarios so i could put in a decent competition.

Okay, so in week one, I found that in the veto, players were  playing mainly to survive, not to be in power. In the end, spaceguy524 fell just short as he didn't respond quick enough in the competition. This went on until the jury opened. By this time you could tell that there was a definite alliance going on and I knew it was between Zapi and Mewto; it was hard to distinguish the others. One of the best cases of teamwork was in one competition where your answer had to be part of the majority to remain in the game. Well, this alliance took up the majority and they all put yes for the answers and insured their alliance won HOH. It was a beautiful move. As the game went on it was mainly between Echinodermata, Zapi, and Mewto. I think the other players got eliminated because they couldnt keep up with all elements of the game.

Lets talk about the individual players.

Zapi2007: You probably would have won four different competitions but you always got bad luck where u were the third to pm me or something like that.

Izzy: You became inactive for the last five eliminations, but nobody ever voted to evict you. Good Job.

Eggy: You almost always came in second during each competition. Congrats on making second.

Meewhee009: What an interesting player, you kind of butchered up a lot of the competitions but you had that alliance going on. Its quite amazing how you got that far into the game.

Echinodermata:A strong player all around.

Mewto55555: Okay. In each competition there was some sort of loophole and you found alot of them. For some reason, you made alot of enemies but you were never evicted. This is how winning competitions gets you ahead.

Okay, i may have gotten these wrong, feel free to discuss this season. I really dont know if I will do something like this again, as it is quite alot of work but it has been fun.
 Thanks guys for playing. Congratulations Mewto55555 your prize is bragging rights.

[quote="Nachosaurus"]Meewhee009: What an interesting player, you kind of butchered up a lot of the competitions but you had that alliance going on. Its quite amazing how you got that far into the game.[/quote]

Hehe... I knew I had an advantage at the start.
I've known you since you were xsquare.  :rotfl:

Totally.
Spaceguy524. Well, you were eliminated week one so i have no words to say


here's summin interesting
so how will our good friend Grobber catch the lion in the desert.?
easy
first he will lock himself in a large spherical cage in the desert.then he will perform a space inversion (inversion freak that he is) with respect to the sphere.by doing this ,he will be outside the cage AND the lion inside it. :D  :D

hey no offence ,grobber
cheers

Hehe nice joke, I haven't seen that for a while.


[quote="warut_suk"]Find the smallest natural number that is divisible by 28, ends in 28 (in base 10 representation) and the sum of  digits is 28.[/quote]

Let this number be $N$. Define $x$ to be an integer such that $N=100x+28$. Since $28|N$ and $28|28$, then $28|100x$ and $7|x$. Furthermore, the sum of the digits of $N$ is just $10$ plus the sum of the digits of $x$, so now we just want the smallest multiple of $7$ that has its digits sum to $18$. Since $x=99$ is not divisible by $7$, we move to the next number with sum of digits $18$, which is $189$. This is divisible by $7$, so we have $N=18928$.


calculus
integration
derivative
linear algebra
matrix
topology
analytic geometry

[quote]Let $ M$ be a compact manifold.  Let $ f: M \rightarrow M$ be a $ C^1$ map such that there exists constants $ K > 0$ and $ \lambda > 1$ such that
\[ K \lambda^n \leq \| df^n(x)\|\]
for all $ x \in M$ and $ n \geq 1$.

Prove that if $ m(A) \equal{} 0$ then $ m(f^{ \minus{} 1}(A)) \equal{} 0$,  where $ m$ denotes the Lebesgue measure.
[/quote]

I don't know how to relate the measure of a set in $ M$ by using the differential of $ f$ if that is the way to prove this  :mad:

Measure is the integral of $ 1$ taken over the set. There is something you know about change of variables in an integral, and it does involve the derivative.

I am not sure about the meaning of $ \|df^n(x)\|$. Is is the derivative of the $ n$th iterate of $ f$ or the $ n$th power of $ df$ understood as a matrix?

Either way, what's with the following example: $ M\equal{}S^1\times S^1$ (a torus), $ f$ wraps the first $ S^1$ twice around itself and shrinks some arc of second $ S^1$ into a point. This seems to be a counterexample, at least to my reading of the question. Anyway, a clarification of $ \|df^n(x)\|$ is needed.

[quote="mlok"]

a clarification of $ \|df^n(x)\|$ is needed.[/quote]

$ \|df(x)\| \equal{} \sup\{\frac{\|df(x)v\|}{\|v\|}  : v \in T_xM, v\neq 0 \}$

[quote="mlok"]Measure is the integral of $ 1$ taken over the set. There is something you know about change of variables in an integral, and it does involve the derivative.

[/quote]

I know the change of variables in $ R^n$, but is the same result true when working with manifolds?

Manifold is covered by coordinate patches, and it suffices to prove that measure is zero within each patch..

In my example with the torus $ f$ has $ \|df\|\ge 2$ everywhere and its $ n$th iterate has $ \|d(f^n)\|\ge 2^n$. Is that a counterexample?

(You did not confirm if $ f^n$ is indeed the $ n$th iteration of $ f$.)


When \[x^5, \quad x+\frac{1}{x}\quad \text{and}\quad 1+\frac{2}{x} + \frac{3}{x^2}\] are multiplied, the product is a polynomial of degree

$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 $

[hide]
C. $6$

Rewrite as $(x^5)(x^1+x^{-1})(x^0+2x^{-1}+3x^{-2})$
It is obvious that the highest term of x that occurs in the product is $x^{5+1} = x^6$
[/hide]

[hide]The $x^5$ term contributes a degree of $5$. it will eventually be multiplied by $x$ (in the $x+1/x$ term) which will make it $6$. it will also be multiplied by $1$ which will make it $6$ still.

[b]C[/b][/hide]

[hide="Answer"]The degree is based on the $x$ with the largest exponent, so $x^5\cdot x\cdot 1=x^6\Rightarrow \boxed{C}$.[/hide]

but part of the definition of a polynomial is that the every exponent is a nonzero integer...so maybe

E=8

[quote="JesusFreak197"]That can't be right, though, because the lowest is $x^{-3}$, so the answer would have to be $9$, which is impossible.

[b]EDIT:[/b] No, nevermind. Remember, every term is multiplied by $x^5$, so there are no negative exponents.[/quote]

wow, i cannot believe that i missed that...so yes, 6 is right


limit
logarithms
calculus
calculus computations

$ \displaystyle\lim_{n\to\infty}{a_n}=a<0$,prove or disprove $ {\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^{a_{n}}}}}$ is convergent

I'm going to assume that you meant $ a>0,$ else it's silly. But even there, it could be false.

For instance, choose $ a_n$ so that $ n^{a_n}\equal{}(2\plus{}(\minus{}1)^n)n.$ That means that $ a_n\equal{}\frac{\ln(2\plus{}(\minus{}1)^n)\plus{}\ln n}{\ln n}$ and $ a_n\to 1.$

But that makes $ n^{a_n}\equal{}\begin{cases}3n&\text{if }n\text{ is even}\\n&\text{if }n\text{ is odd}\end{cases}.$ That makes our series

$ \minus{}1\plus{}\frac16\minus{}\frac13\plus{}\frac1{12}\minus{}\frac15\plus{}\frac{1}{18}\plus{}\cdots\minus{}\frac1{2k\minus{}1}\plus{}\frac1{3k}\plus{}\cdots$

and that diverges to $ \minus{}\infty.$


inequalities
real analysis
real analysis unsolved

Let $ A\subset \mathbb{R}^{\plus{}}$, $ B\subset \mathbb{R}^{\plus{}}$ non empty sets.

Let $ A\cdot B: \equal{}\{x\cdot y: x\in A\wedge y\in B\}$. Prove that:

i) If $ A$ and $ B$ are bounded $ \implies A\cdot B$ is bounded ( sup and inf )

ii) $ \sup A\cdot B\equal{}\sup A \cdot \sup B$ and $ \inf A\cdot B \equal{}\inf A \cdot \inf B$

i) let A and B be bounded, then

$ \exists a_1,a_2$ s.t. $ a_1 \le a \le a_2, \forall a \in A$
$ \exists b_1,b_2$ s.t. $ b_1 \le b \le b_2, \forall b \in B$

let $ x\equal{}ab \in A\cdot B$
then $ a_1b_1\le ab_1\le ab\le a_2b \le a_2b_2$ so $ A\cdot B$ is bounded

ii)

let $ |\sup A|\plus{}|\sup B|\equal{}c>>\epsilon > 0$

then $ \forall a\in A, b\in B, ab\le \sup A \sup B \to \sup AB \le \sup A \sup B$

$ \exists a_o \in A$ s.t. $ a_o > \sup A \minus{} \left(\frac{c }{2}\minus{}\sqrt{\frac{c }{2}\minus{}\epsilon}\right)$
$ \exists b_o \in B$ s.t. $ b_o > \sup B \minus{} \left(\frac{c }{2}\minus{}\sqrt{\frac{c }{2}\minus{}\epsilon}\right)$

then $ a_ob_o >(\sup A \minus{} \left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right) ) (\sup B \minus{} \left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right))$
$ \equal{} \sup A \sup B \minus{} (\sup A \plus{} \sup B )\left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right) \plus{} \left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right)^2$
$ \ge \sup A \sup B \minus{} c\left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right)\plus{}\left(\frac{c }{2}\minus{}\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\right)^2$
$ \equal{}\sup A \sup B \minus{} \frac{c^2}{2} \plus{}c\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\plus{}\frac{c^2}{4}\minus{}c\sqrt{\frac{c^2 }{4}\minus{}\epsilon}\plus{}\frac{c^2 }{4}\minus{}\epsilon$
$ \equal{}\sup A \sup B \minus{} \epsilon$  we gain the inequality

$ \sup AB \ge \sup A \sup B \minus{}\epsilon$ and as epsilon goes to zero we get the equality desired.

inf is done the same.


The roots of x2+ax+1 = b are positive integers. Prove that the integer a2 + b2 is composite.

[hide]The quadratic is x^2 + ax + (1-b) = 0. Let the integers roots be r and s. We have -(r+s) = a and rs = 1-b <=> b = 1-rs. Substituting, we find that

a^2 + b^2 = (-(r+s))^2 + (1-rs)^2

= r^2 + 2rs + s^2 + 1 + (rs)^2 - 2rs

= r^2 +  s^2 + 1 + (rs)^2

= (r^2+1)(s^2+1)

Ta-da![/hide]

good job, just a minor technicality though.  If you see this type of problem on an olympiad or other contest, make sure you explicitly say that neither of the factors (r^2+1) nor (s^2+1) is equal to 1 or -1--which is trivial in this case-- because primes do factor as 1*p.

err...do you mean "neither the factor (r^2+1) nor the factor (s^2+1) is equal to 1"?  I think there was an extra not in there, which made the sentence mean "both of the factors are equal to 1".

believe it or not, I know what a double negative is, I just typed it out quickly.  I'm pretty sure eveyone understood what i meant to say.


Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $ n$ to $ 5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the number of beans Alice has is greater than $ 2008$ and has last two digits $ 42$. What is the minimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

[hide="hope this is right."]This seems a really bad idea but its more logical than mathematical (im sure this isnt the intended solution)

but obviously we want to expand by as much as possible in the earlier turns to get bigger than 2008

so 1-5-25-125-625.

Now heres were the logic comes in, if you do 625*5, you get 3125, which is bigger than 2008 but very hard to get to 3142 without having to pay a lot of money.  Also, notice that whenever you multiply a power of 5 to 5, your last two digits will always be 25 so theres no point in keep on multiplying by 5.

So from the 625, we want to add  some 1s so that when we multiply, the answer's last two digits will be very close to 42.  Clearly we want the last two digits to be 040, so 625-626-627-628.  Then 628-3140-3141-3142.

That is 5+1+1+1+1+5+5+5+1+5+5=30+5=35 cents.

Also notice that you want to go to the 28 ending as soon as  possible (while keeping it when multiplied by 5 over 2008) to save money, theres no need to do 625*5, then add 3 1s to get to the 28 ending because that will just be wasting 1 cent.[/hide]

[hide="Solution"]
We show an algorithm which requires only $ 35$ cents:
\[ \begin{tabular}{|r\parallel{}r|r|r|r|r|r|r|r|r|r|r|r|}
\hline
\text{Cents}&0&5&6&7&8&9&14&19&24&25&30&35\\
\hline
\text{Beans}&0&1&5&25&125&625&626&627&628&3140&3141&3142\\
\hline
\end{tabular}\]
Now suppose we could achieve this with less than $ 35$ cents. Since $ 42 \equiv 2 \pmod{5}$, and the first option makes the number of beans in the pile $ \equiv 0 \pmod{5}$, it follows that we must use the second option at least twice at the end. If we used the second option more than twice, that it must be used at least $ 7$ times, which would by itself cost $ 7 \times 5 \equal{} 35$ cents, contradiction. So we must get our pile to reach $ \equiv 40 \pmod{100}$ and then use the second option twice at a cost of $ 10$ cents. Backtracking another move, we have to use the first option (for a total cost of $ 11$ cents), and we need to get our pile to reach $ \equiv 8 \pmod{20}$.

By the same logic, we need to use the second option three times again, for a total cost of $ 26$ cents. Also, since we started with $ 0$ beans, we must've used the second option for the first move, and our total is now at least $ 31$ cents.

Finally, we need to satisfy the condition that the number of beans is $ > 2008$. This clearly requires at least 5 quituplings, of which one was done already. Hence, the minimal amount must be at least $ \boxed{35}$, for which we showed an example that works.
[/hide]

Why is my solution wrong :o
Notice that in base 5, giving a penny adds a tailing 0 and giving a nickel adds a 1. Also, the sequence must be $>2018=11013_5.$ Let $m$ be the minimum ending sequence. Then, we have

$$(100 \cdot \overline{a_1a_2a_3...a_i}+42)_{10} = (\overline{b_1b_2b_3...b_j}+132)_5$$
since $42_{10} = 132_5.$

Now, subtracting $42_{10}$ from both sides and noticing $100_{10} = 400_5,$ we see that

$$400_5 | (\overline{b_1b_2b_3...b_j})_5,$$
or the last two digits after subtraction is $00$ and $b_1+b_2+b_3+...+b_j = 0 (mod 4). $

Using this, we construct a sequence.

$$0,1,10,11,110,111,1110,1111,11110,11111, 111110, 111111, 111112, 111113, 1111130, 1111131, 1111132.$$
This satisfies all constraints, so we are done

Edit: I see why this is wrong, but the official solution of getting 1000_10 is quite... out of the blue :(


[quote="Los"]How would you find 1680\mod{2771} without using fancy congruence tricks? Can you use Euclid's algorithm in some way?[/quote]

You don't need any trick at all, $ 1680 < 2771$, therefore $ 1680 \equiv 1680 (\mbox{mod } 2771)$

hm. how would you find $ 1680^2\plus{}1\pmod{2771}$? (this is \equiv 1523)?

you use division... I suppose if you wanted to lessen the computation, you could use 1680 is the same as 1680-2771...


function
floor function
algebra proposed
algebra

$ (\forall)a\in \mathbb{R}$ , prove that :
\[ i) \left\lfloor a \plus{} \frac {1}{2}\right\rfloor \plus{} \left\lfloor a \plus{} 1\right\rfloor \plus{} \left\lfloor 2a \plus{} \frac {1}{2}\right\rfloor \equal{} \left\lfloor 4a \plus{} 1\right\rfloor\]

\[ ii)\left\lfloor a \plus{} \frac {1}{6}\right\rfloor \plus{} \left\lfloor a \plus{} \frac {7}{12}\right\rfloor \plus{} \left\lfloor a \plus{} \frac {5}{6}\right\rfloor \plus{} \left\lfloor 3a \plus{} \frac {1}{4}\right\rfloor \equal{} \left\lfloor 4a \plus{} \frac {1}{3}\right\rfloor \plus{} \left\lfloor a \plus{} \frac {5}{12}\right\rfloor \plus{} \left\lfloor a \plus{} \frac {3}{4}\right\rfloor\]

$ i).$ Let $ a\equal{}x\plus{}\alpha$

(*)
  $ 0 <\alpha <\frac{1}{2}$

Then $ \left\lfloor a\plus{}\frac{1}{2} \right\rfloor  \plus{}\left\lfloor a \plus{}1 \right\rfloor \plus{}\left\lfloor 2a \plus{}\frac{1}{2} \right\rfloor \equal{}x \plus{}x\plus{}1 \plus{}2x\equal{}\left \lfloor 4a \plus{}1 \right \rfloor \equal{}4x\plus{}1$

(**)

$ \frac{1}{4} <\alpha <\frac{1}{2}$

Then $ \left \lfloor a\plus{}\frac{1}{2} \right \rfloor \plus{} \left \lfloor a\plus{}1 \right \rfloor  \plus{}\left \lfloor 2a \plus{}\frac{1}{2} \right \rfloor \equal{}x\plus{}x\plus{}1 \plus{}2x\plus{}1 \equal{}\left \lfloor 4a \plus{}1 \right \rfloor \equal{}4x\plus{}2$

(***)

$ \frac{1}{2} <\alpha <\frac{3}{4}$

Then $ \left \lfloor a\plus{}\frac{1}{2} \right \rfloor \plus{} \left \lfloor a\plus{}1 \right \rfloor  \plus{}\left \lfloor 2a \plus{}\frac{1}{2} \right \rfloor \equal{}x\plus{}1\plus{}x\plus{}1 \plus{}2x\plus{}1 \equal{}\left \lfloor 4a \plus{}1 \right \rfloor \equal{}4x\plus{}3$

(****)

$ \frac{3}{4} < \alpha <1$

Then $ \left \lfloor a\plus{}\frac{1}{2} \right \rfloor \plus{} \left \lfloor a\plus{}1 \right \rfloor  \plus{}\left \lfloor 2a \plus{}\frac{1}{2} \right \rfloor \equal{}x\plus{}1\plus{}x\plus{}1 \plus{}2x\plus{}2 \equal{}\left \lfloor 4a \plus{}1 \right \rfloor \equal{}4x\plus{}4$


geometry
trapezoid
angle bisector
geometry unsolved

In a circumference with center $ O$ we draw two equal chord $ AB\equal{}CD$ and if $ AB \cap CD \equal{}L$ then $ AL>BL$ and $ DL>CL$
We consider $ M \in AL$ and $ N \in DL$ such that $ \widehat {ALC} \equal{}2 \widehat {MON}$
Prove that the chord determined by extending $ MN$ has the same as length as both $ AB$ and $ CD$

if possible can you draw it to see clearly the exercise
thank you very much. :)

Let $OM$ and $ON$ cut line segment $BC$ at $E$ and $F$ respectively.And let $P$ and $Q$ be projections of $O$ to $AB$ and $CD$ respectively.$AB=CD$ and $ABCD$-cyclic implies that $ABCD$ is isosceles trapezoid.$BL=CL \implies AL=DL \implies \angle BAD=\angle CDA = \angle DCB = \angle ABC$. $OP \perp AB$ and $OQ \perp CD$ implies that $AP=PB$ and $QD=QC;AL=DL$ implies that $PL=QL \implies \angle LPQ = \angle LQP= \angle ABC$ which implies that $BC || PQ ;OL \perp PQ \implies \angle MOP= \angle NOL$ and $OL \perp BC \implies \angle DCB = \angle NOE \implies CENO$-cyclic and similarly we get that $BFMO$ is cyclic and from $OB=OC$ we get that $\angle OBC= \angle OCB \implies \angle OCN =\angle OBM=\angle OEN=\angle OFM$ which implies that $MEFN$ is cyclic $\implies \angle OFE=\angle OMN$ and $\angle MOP=\angle NOL$ implies that $\angle OFE=\angle OMP \implies \angle OMP=\angle OMN \implies \angle OMX= \angle OMB$ if $MN$ cuts circle at $X$ and $Y$. Let angle bisector of $\angle BOX$ cut $XM$ at $K$ and let $WLOG$ $K$ lies between
$X$ and $M \implies \angle OMX+\angle KOM =\angle OKX=\angle OKB$
$\implies \angle OKB >\angle OMX=\angle OMB$ which is contradiction thus we get that $M=K \implies MB=MX \implies AXBY$ is isosceles trapezoid which implies that $XY=AB$.
_____________________________________________________
[i]by Elshad Mustafayev[Sumgayit][/i]

It's enough to prove that $O$ is center of the excircle of $LMN$ and this is obvious because $2MON=ALC$ and $DLO=OLA$


inequalities
trigonometry
inequalities proposed

Let $ x,y,z > 0, x \plus{} y \plus{} z \equal{} \pi$ prove that
\[ 1 \plus{} \sin\frac {x}{2}\sin\frac {y}{2}\sin\frac {z}{2}\ge\sqrt {3}(\sin x\sin y\sin z)
\]

Easy and well-known:
$ \prod_{x,y,z}\sin{\frac {x}{2}} \equal{} \frac {r}{4R}$.
$ \prod_{x,y,z}\sin{x} \equal{} \frac {S}{2R^2}$.
So we must prove that
$ 2R^2 \plus{} \frac {Rr}{2}\geq\sqrt {3}S$,ro
$ 4R^2 \plus{} Rr\geq\sqrt {12}S$.
or
$ R\frac {(4R \plus{} r)}{S}\geq\sqrt {12}$.
It is easy to understand that
$ \frac {4R \plus{} r}{S} \equal{} \sum_{a,b,c}\frac {1}{p \minus{} a}$,where $ a,b,c$ are the sides length of the triangle with angles $ x,y,z$ and
$ p$ is equal to $ \frac {a \plus{} b \plus{} c}{2}$.
so we must prove that
$ \sum_{a,b,c}\frac {R}{p \minus{} a}\geq\sqrt {12}$.
Now if we use CBS inequality we'll obtain:
$ \sum_{a,b,c}\frac {1}{p \minus{} a}\geq\frac {9}{p}$.
So we must prove that
$ \frac {R}{p}\geq\frac {\sqrt {12}}{9}$.
or which is equivalent to this one:
$ \frac{p}{4R}\leq\frac{3\sqrt {3}}{8}$.
But $ \frac {p}{4R} \equal{} \prod_{x,y,z}\cos{\frac {x}{2}}$.
So our inequality is equivalent to:
$ \prod_{x,y,z}\cos{\frac {x}{2}}\leq\frac {3\sqrt {3}}{8}$.
Which is well-known and not hard to prove.

Thanks, my solution is the same but
\[ 4R\frac {(4R \plus{} r)}{S}\geq\sqrt {12}\]
it is weaker than
$ r(4R\plus{}r)\geq r\sqrt{3}s\geq \sqrt{3}S$ :)


geometry
function
inequalities
perimeter
algebra
domain
area of a triangle

Suppose $ a ,b,c$ are three sides of $ \triangle ABC$ and $ s\equal{}\frac{a\plus{}b\plus{}c}{2}.$

Given that the area of $ \triangle ABC$ is $ 1$.

Find the possible least value of $ (s\minus{}a)^3\plus{}(s\minus{}b)^3\plus{}(s\minus{}c)^3$

 :D  :lol:

[quote="PKMathew"]Suppose $ a ,b,c$ are three sides of $ \triangle ABC$ and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}.$

Given that the area of $ \triangle ABC$ is $ 1$.

Find the possible least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3$

 :D  :lol:[/quote]
[hide="This may be a solution :"]

Since $ (s \minus{} a), (s \minus{} b), (s \minus{} c)$ are positive terms.

Use AM_GM $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3\sqrt[3] {(s \minus{} a)^3  (s \minus{} b)^3  (s \minus{} c)^3}$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3(s \minus{} a)(s \minus{} b) (s \minus{} c)$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac{{3s(s \minus{} a)(s \minus{} b) (s \minus{} c) }}{s}$
$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac{3\times1^2}{s}\equal{}\frac{6}{a\plus{}b\plus{}c}$

The eqaulity holds when  $ (s \minus{} a)\equal{}  (s \minus{} b)\equal{}  (s \minus{} c) \iff a\equal{}b\equal{}c$

If $ a\equal{}b\equal{}c, \triangle ABC$ is equilateral so that  $ \frac{1}{2} a^2 \sin60^\circ\equal{}1 \implies a\equal{}\frac{2}{\sqrt[4]{3}}$

$ \therefore$ the least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\equal{}\frac{2}{a}\equal{}\sqrt[4]{3}$

[/hide]:juggle:

[quote="vinskman"][quote="PKMathew"]Suppose $ a ,b,c$ are three sides of $ \triangle ABC$ and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}.$

Given that the area of $ \triangle ABC$ is $ 1$.

Find the possible least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3$

 :D  :lol:[/quote]
[hide="This may be a solution :"]

Since $ (s \minus{} a), (s \minus{} b), (s \minus{} c)$ are positive terms.

Use AM_GM $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3\sqrt [3] {(s \minus{} a)^3 (s \minus{} b)^3 (s \minus{} c)^3}$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3(s \minus{} a)(s \minus{} b) (s \minus{} c)$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac {{3s(s \minus{} a)(s \minus{} b) (s \minus{} c) }}{s}$
$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac {3\times1^2}{s} \equal{} \frac {6}{a \plus{} b \plus{} c}$

The eqaulity holds when  $ (s \minus{} a) \equal{} (s \minus{} b) \equal{} (s \minus{} c) \iff a \equal{} b \equal{} c$

If $ a \equal{} b \equal{} c, \triangle ABC$ is equilateral so that  $ \frac {1}{2} a^2 \sin60^\circ \equal{} 1 \implies a \equal{} \frac {2}{\sqrt [4]{3}}$

$ \therefore$ the least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3 \equal{} \frac {2}{a} \equal{} \sqrt [4]{3}$

[/hide]:juggle:[/quote]

Nice.
But I have one question>
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \ge n\sqrt [n]{a_1 a_2 a_3 \cdots a_n}..... a_i \ge 0$
Is the min value of $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$ when $ a_1 \equal{} a_2 \equal{} a_3 \equal{} \ldots \equal{} a_n$ ??
just a small question  :blush:

Actually, $ a_1 \plus{} a_2 \plus{}\cdots \plus{} a_n \ge n\sqrt[n]{a_1 a_2 a_3 \cdots a_n}$ And the min value is $ a_1\equal{}a_2\equal{}a_3\equal{}a_n$ that make the $ LHS\equal{}RHS$.

[quote="doabli"]Actually, $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \ge n\sqrt [n]{a_1 a_2 a_3 \cdots a_n}$ And the min value is $ a_1 \equal{} a_2 \equal{} a_3 \equal{} a_n$ that make the $ LHS \equal{} RHS$.[/quote]
Sorry ... I edited it  :blush: ,Thank for info.  :D

[quote="Altheman"]vinskman, your solution isn't correct. you proved that the function in question is greater than or equal to 6/(a+b+c), but you did not find the minimum of that expression.

Soln.

Let $ x \equal{} s \minus{} a$, a.s.o.

Then we want to minimize $ x^3 \plus{} y^3 \plus{} z^3$ subject to the condition that $ xyz(x \plus{} y \plus{} z) \equal{} 1$.

Observe that
\[ a(x^3 \plus{} y^3 \plus{} z^3)\ge \frac {a}{9}\cdot (x \plus{} y \plus{} z)^3\qquad (1)
\]

\[ b(x^3 \plus{} y^3 \plus{} z^3)\ge 3bxyz\qquad (2)
\]

\[ b(xyz \plus{} xyz \plus{} xyz) \plus{} \frac {a}{9}\cdot (x \plus{} y \plus{} z)^3\ge \sqrt [4]{b^3x^3y^3z^3\frac {a}{9}(x \plus{} y \plus{} z)^3} \equal{} 4\frac {\sqrt [4]{ab^3}}{\sqrt {3}} \qquad (3)
\]
In (1) and (2), equality occurs when $ x \equal{} y \equal{} z$ so by the given condition, $ 3x^4 \equal{} 1$, so $ x \equal{} \frac {1}{\sqrt [4]{3}}$. For equality in (3), we need $ bxyz \equal{} \frac {a}{9}(x \plus{} y \plus{} z)^3$ so $ b \equal{} 3a$. Adding (1), (2), and (3), we get
\[ (a \plus{} b)(x^3 \plus{} y^3 \plus{} z^3)\ge \frac {4\sqrt [4]{ab^3}}{\sqrt {3}}\qquad (4)
\]
We take $ a \plus{} b \equal{} 1$, so $ a \equal{} \frac {1}{4}$ and $ b \equal{} \frac {3}{4}$. The minimum is then $ \sqrt [4]{3}$ which occurs for $ x \equal{} y \equal{} z \equal{} \frac {1}{\sqrt [4]{3}}$.[/quote]

Sorry,
Haha, your works are not wiser than mine..

as doabli's explain: Actually, $ a_1 \plus{} a_2 \plus{}\cdots \plus{} a_n \ge n\sqrt[n]{a_1 a_2 a_3 \cdots a_n}$And the min value is  $ a_1\equal{}a_2\equal{}\cdots\equal{}a_n$that make the$ LS\equal{}RS$ .

[quote="vinskman"][quote="Altheman"]vinskman, your solution isn't correct. you proved that the function in question is greater than or equal to 6/(a+b+c), but you did not find the minimum of that expression.
[/quote]

Sorry,
Haha, your works are not wiser than mine..

as doabli's explain: Actually, $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \ge n\sqrt [n]{a_1 a_2 a_3 \cdots a_n}$And the min value is  $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n$that make the$ LS \equal{} RS$ .[/quote]

Not quite; you're bounding the quantity $ f(a,b,c)$ above a non-constant function, but that non-constant function can certainly take values lower than those achieved at equality.

For example, if I choose $ (a,b,c)\equal{}(13/\sqrt{60}, 13/\sqrt{60}, 24/\sqrt{60})$, the inequality you give yields $ f(a,b,c) > \frac{6}{a\plus{}b\plus{}c} \equal{} \frac{3\sqrt{60}}{25} \approx 0.93$, while the correct lower bound is at $ \approx 1.32$.  :wink:

[quote="vinskman"][quote="PKMathew"]Suppose $ a ,b,c$ are three sides of $ \triangle ABC$ and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}.$

Given that the area of $ \triangle ABC$ is $ 1$.

Find the possible least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3$

 :D  :lol:[/quote]
[hide="This may be a solution :"]

Since $ (s \minus{} a), (s \minus{} b), (s \minus{} c)$ are positive terms.

Use AM_GM $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3\sqrt [3] {(s \minus{} a)^3 (s \minus{} b)^3 (s \minus{} c)^3}$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge 3(s \minus{} a)(s \minus{} b) (s \minus{} c)$

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac {{3s(s \minus{} a)(s \minus{} b) (s \minus{} c) }}{s}$
$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac {3\times1^2}{s} \equal{} \frac {6}{a \plus{} b \plus{} c}$

The eqaulity holds when  $ (s \minus{} a) \equal{} (s \minus{} b) \equal{} (s \minus{} c) \iff a \equal{} b \equal{} c$

If $ a \equal{} b \equal{} c, \triangle ABC$ is equilateral so that  $ \frac {1}{2} a^2 \sin60^\circ \equal{} 1 \implies a \equal{} \frac {2}{\sqrt [4]{3}}$

$ \therefore$ the least value of $ (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3 \equal{} \frac {2}{a} \equal{} \sqrt [4]{3}$

[/hide]:juggle:[/quote]

Yeah, my solution's the same as yours.

By applying Heron's Formula we can easily obtain the inequality, and then we make sure it is an equilateral
triangle, and then we find the minimum value.

Actually vinskman, your proof have a problem right here, like Altheman said.
[quote="vinskman"]

$ \implies (s \minus{} a)^3 \plus{} (s \minus{} b)^3 \plus{} (s \minus{} c)^3\ge \frac {3\times1^2}{s} \equal{} \frac {6}{a \plus{} b \plus{} c}$

[/quote]

$ \frac{6}{a\plus{}b\plus{}c}$ is not a bound because it has a,b,c (they are all variable$ \implies$ they can change the minimum value). A bound need to be a constant. not full of variable like that.
And, can anyone explain for me the (4) of Altheman's proof, I don't understand why we take $ a\plus{}b\equal{}1$
PS: Altheman, your (3). you have a mistake in here
$ b(xyz \plus{} xyz \plus{} xyz) \plus{} \frac {a}{9}\cdot (x \plus{} y \plus{} z)^3\ge \sqrt [4]{b^3x^3y^3z^3\frac {a}{9}(x \plus{} y \plus{} z)^3} \equal{} 4\frac {\sqrt [4]{ab^3}}{\sqrt {3}} \qquad (3)$
It suppose to be like this
$ b(xyz \plus{} xyz \plus{} xyz) \plus{} \frac {a}{9}\cdot (x \plus{} y \plus{} z)^3\ge 4\sqrt [4]{b^3x^3y^3z^3\frac {a}{9}(x \plus{} y \plus{} z)^3} \equal{} 4\frac {\sqrt [4]{ab^3}}{\sqrt {3}} \qquad (3)$

I think Altheman meant to take $ a,b$ as weights (though ambiguous because $ a,b$ were already defined in the problem). He was breaking the $ x^3 \plus{} y^3 \plus{} z^3$ expression into $ a(x^3 \plus{} y^3 \plus{} z^3) \plus{} b(x^3 \plus{} y^3 \plus{} z^3)$ and applying different inequalities to both parts in a way that was convenient. So that's why $ a \plus{} b \equal{} 1$.

vinskman, your solution is wrong. Take a triangle with area 1 and perimeter larger than an equilateral triangle; the bound $ \frac{6}{a\plus{}b\plus{}c}$ that you've given is now lower than the minimum you claim. How do you know no such triangle can bring the quantity lower?

Thank you, I see, your proof is great Altheman.

Query for Altheman:
We have to minimize $ x^3\plus{}y^3\plus{}z^3$ under the constraint $ xyz(x\plus{}y\plus{}z) \equal{} 1$

Can we not use Lagrange Multipliers here as it directly gives that $ x\equal{}y\equal{}z \equal{} \frac{1}{\sqrt [4] 3}$ ?

Not directly... The domain is unbounded. i.e. there is no constant M such that $ \sqrt{x^2\plus{}y^2\plus{}z^2}<M$ for all $ x,y,z$ such that $ xyz(x\plus{}y\plus{}z)\equal{}1$. You could try to argue that the minimum has to occur when $ x^2\plus{}y^2\plus{}z^2<M$...


algebra
polynomial
calculus
calculus computations

Let $ f(x)$ is polynomial solve $ f(x)$ from $ f(2x)\equal{}f'(x)f''(x)$  :fool:

I have $ f(x) \equal{} \frac {4}{9}x^3,0$. Who are confirm ??
ps. is 0 polynomial ??  :(

Yes, those are the only solutions.  (The zero polynomial is certainly a polynomial.)

There's a portion of the forum devoted to calculus, and this is not it.  In the future, if you have calculus problems, you can put them [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]here[/url].  (Don't double-post this problem, though -- hopefully a moderator will move it.)

take : $ f(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{} . . . \plus{}a_{1}x\plus{}a_{0}$

then it's easy :wink:


Find all pairs of positive integers $ (n,k)$ such that k>1 and $ A\equal{}17^{2006n}\plus{}4.17^{2n}\plus{}7.19^{5n}$ is the product of k integers in succession. (Ex: 3,4,5,6 are four integers in succession)

To [b]mszew[/b]

How will you show that for all such n,$ A$ can be expressed as a product of k consecutive integers??

There just necessary conditions... a very small step towards the solution...
besides the trivial ones with k=1

3 days ago, I found one solution for the problem.
It is easy to show that k<4.
$ n \equal{} 0, A \equal{} 6 \equal{} 2.3$
Examine the problem when $ n\geq 1$ (because we must find all pairs of positive integers $ (n,k)$ )
* With k=2, $ (17^{1003n})^2 < A \equal{} m(m \plus{} 1) < (17^{1003n} \plus{} 1)^2$ when $ n\geq 1$.
So, $ m < 17^{1003n} \plus{} 1$ and $ m \plus{} 1 > 17^{1003n}$. It shows that $ m \equal{} 17^{1003n}$.
$ A \equal{} 17^{1003n}.(17^{1003n} \plus{} 1) (*)$ . It's easy show that (*) isn't true.
* With k=3, by the proof of [b]mszew[/b] , we have $ A\equiv 2, 8 (mod5)$.
If a product of 3 consecutive integers isn't divisible by 5, it will $ \equiv 1.2.3\equiv 6$ or $ \equiv 2.3.4\equiv 4 (mod 5)$. The statement show that $ A\equiv 0,6,4 (mod5)$.
Then, A can't be expressed as a product of 3 consecutive integers.
No solution.


geometry
geometric transformation
reflection
analytic geometry

A cubical box with sides of length 7 has vertices at (0,0,0), (7,0,0), (0,7,0), (7,7,0), (0,0,7), (7,0,7), (0,7,7), (7,7,7). The inside of the box is lined with mirrors and from the point (0,1,2), a beam of light is directed to the point (1,3,4). The light then reflects repeatedly off the mirrors on the inside
of the box. Determine how far the beam of light travels before it first returns
to its starting point at (0,1,2).

[hide]$ \vec{v_{i}}\equal{}\frac{\vec{p_{1}}\minus{}\vec{p_{0}}}{\Delta t}\equal{}\langle 1,3,4\rangle\minus{}\langle 0,1,2\rangle \equal{}\langle 1,2,2\rangle$

We can consider the three orthagonal directions independently:

In the $ x$ direction, it makes a full lap (back and forth) every 14 seconds, and only achieves it's $ x$ coordinate once per lap (since it is at an extreme end of the lap). So, it will return to it's $ x$ coordinate at times of the form $ t \equal{} 14k$ for positive integers $ k$.

In the $ y$ direction, it makes a full lap every 7 seconds, and will will first cross in the backwards direction at a time of $ \frac{2(7\minus{}1)}{2}\equal{} 6$. So, it will return to its original $ y$ coordinate at $ t \equal{} 7k$ or at $ t \equal{} 7k\plus{}6$

In the $ z$ direction, it makes a full lap every 7 seconds, and will will first cross in the backwards direction at a time of $ \frac{2(7\minus{}2)}{2}\equal{} 5$. So, it will return to its original $ z$ coordinate at $ t \equal{} 7k$ or at $ t \equal{} 7k\plus{}5$.

The umbrella solution for the above three criteria is $ t \equal{} 14k$ (since, by the first, it must be divisible by 7), and our first positive solution is at $ k \equal{} 1\implies t \equal{} 14$.

$ d \equal{}\|\vec{v_{i}}\|\Delta t \equal{}\left(\sqrt{1^{2}\plus{}2^{2}\plus{}2^{2}}\right) (14) \equal{}\boxed{42}$[/hide]


[quote="stvs_f"]let a,b,c are positive.prove that:
\[ \sum_{a,b,c} (\frac {a^3}{b^2 \minus{} bc \plus{} c^2})\geq a \plus{} b \plus{} c
\]
[/quote]

$ \leftrightarrow \sum \frac{a^3}{b^3\plus{}c^3\plus{}a(b^2\minus{}bc\plus{}c^2)} \ge 1$
by cauchy,
$ LHS(\sum a(b^3\plus{}c^3\plus{}a(b^2\minus{}bc\plus{}c^2)))\equal{}LHS(\sum_{sym}a^3b\plus{} \sum_{sym}a^2b^2\minus{} \sum_{cyc}a^2bc)\ge(a^2\plus{}b^2\plus{}c^2)^2$
We need to prove $ (a^2\plus{}b^2\plus{}c^2)^2\ge(\sum_{sym}a^3b\plus{} \sum_{sym}a^2b^2\minus{} \sum_{cyc}a^2bc)$.
$ \leftrightarrow \sum a^2(a\minus{}b)(a\minus{}c) \ge 0$, It is obvious.


Let $ m,n\in \mathbb{N}$ such that $ m\mid n$ and $ p$ a prime. Prove that $ p^m\minus{}1\mid p^n\minus{}1$ if and only if $ m\mid n$.

[quote="MeKnowsNothing"]Let $ m,n\in \mathbb{N}$ such that $ m\mid n$ and $ p$ a prime. Prove that $ p^m \minus{} 1\mid p^n \minus{} 1$ if and only if $ m\mid n$.[/quote]
P is not necessary a prime .
$ x\in N$ and $ x>1$ then $ \gcd(x^m\minus{}1,x^n\minus{}1)\equal{}x^{\gcd(m,n)}\minus{}1$
Your problem is a special case of this problem .

Thanks for replies. I was thinking a proof like

Let $ F$ be a field, $ d,n\in \mathbb{N}$, $ x^d\minus{}1,x^n\minus{}1\in K[x]$. Then $ x^d\minus{}1\mid x^n\minus{}1$ iff $ d\mid n$. Then we take the mapping $ g: \mathbb{F}_p\to \mathbb{N}: g(x^{p^n\minus{}1})\mapsto p^n$. But it looks like this won't work because $ \mathbb{N}$ is not an algebra thus $ g$ is not a homomorphism.

It's a little simpler than that and I think you mean $ m < n$.  TTsphn's lemma can be proven as follows:

Let $ d \equal{} \gcd(x^m \minus{} 1, x^n \minus{} 1)$.  Then $ x^m \equiv x^n \equiv 1 \bmod d$.  There exist integers $ r, s$ such that $ mr \plus{} ns \equal{} \gcd(m, n) \implies x^{\gcd(m,n)} \equiv 1 \bmod d$, so $ d | x^{\gcd(m,n)} \minus{} 1$.  And of course $ x^{\gcd(m,n)} \minus{} 1 | d$.


[quote="brady00n"]Hello, I'd like to go back to the basics and really understand all of the logic.  I'm looking for what people generally recognize as the "best" books in each field, which to me means easy to understand the basic points, yet provides challenging problems.
1) Mathematical Logic
2) Discrete Math[/quote]

There's so many books, just search amazon.

One good intro to both discrete math and mathematical logic is Kenneth Rosen's [i]Discrete Mathematics and its Applications.[/i]
The publisher churns the editions---there is almost no difference between editions, so get an old used one for 1/5 the price.
These are not the deepest books on the subject, and they'll only get you up to freshman college level, but they are good starters for an applied view of the field.

If you want a deeper, more "pure" mathematical approach, I don't have any good suggestions---it's been too long since my math-major days and I don't remember any of my logic or abstract algebra books except Hughes and Cresswell [i]Modal Logic[/i] and Serge Lang's [i]Algebra[/i], neither of which is suitable as an introduction to the subjects.


Fie $f:[0,1] \to [0,1]$ o functie continua astfel incat $f( \frac12) \ne \frac12$.
Aratati ca exista $a,b \in (0,1)$ cu proprietatile:$f^2(a)+f^2(b) \ge 2a(1-b).$

eu iti sugerez sa te mai uiti o data pe sursa problemei si dupa aia sa ne-o scrii asa cum trebuie.

Ai dreptate intr-un fel pt ca si eu m-am mirat cand am vazut-o si mai ales cand la indicatie dadea ceva pe langa subiect,deci este gresita scrisa in cartea din care am luat-o. :),am vrut sa vad numai daca chiar este gresita sau nu inteleg eu indicatia.


Let $f$ a fuction $f: R\to R$ sastifying $\large |f(a)-f(b)|<|a-b| \forall a,b \in R; a\neq b$. Prove that if $\large f(f(f(\alpha)))=\beta$ and  $\large f(f(f(\beta)))=\alpha$, we have $\large f(\alpha)=\beta$

$\large f(f(f(\alpha)))=\beta$ and  $\large f(f(f(\beta)))=\alpha$ $\Longrightarrow \alpha=\beta$.

:lol:    Yes; just to be a little less minimalist:

$\alpha\neq\beta\rightarrow \ |\alpha-\beta|=|f(f(f(\alpha)))-f(f(f(\beta)))|<$
$<|f(f(\alpha))-f(f(\beta))|<|f(\alpha)-f(\beta)|<|\alpha-\beta|$

And therefore, holds what Rust said.
On the other hand:
$f(\alpha)\neq\alpha \rightarrow f(f(\alpha))\neq\alpha \rightarrow$
$|\alpha-f(\alpha)|=|f(f(f(\alpha)))-f(\alpha))|<|f(f(\alpha))-\alpha|$

$f(\alpha)\neq\alpha \rightarrow f(\alpha)=f(f(\alpha))$  so

$f(\alpha)\neq\alpha \rightarrow |\alpha-f(f(\alpha))|=|f(f(f(\alpha)))-f(f(\alpha))|<$
$<|f(f(\alpha))-f(\alpha)|<|f(\alpha)-\alpha|$

Which contraddicts  the previous inequality.


ARML
number theory unsolved
number theory

Beeter now?? im soory!! my mistake :lol:

[quote="Γιώργος"]Find the positive nteger $ n$ such that $ 1/n$ is closest to {$ {\sqrt {123456789}}$}[/quote]

[quote="pleurestique"]you sure there isn't something wrong with this problem?[/quote]

I think he is regarding the fact that $ \sqrt {123456789}>1$, hence the best integer will be $ n\equal{}1$


calculus
geometry
inradius
geometric transformation
reflection

hi guys

how many of u have submitted the applications

and how r u planning to go abt it

plz reply and make it active very quickly

thankx

Have you already submitted it?when is the last date? have you submitted Application form
online?

[quote="mittal"]this thread reminded me of the golden olden days :P  :oops:[/quote]
Yeah, when I didn't even know what KVPY was and walked in to give the exam coz of my sir :P Anyways, good luck to all the applicants this year :)

hey murgi hi

yeah i have submitted the form but dont know whether they received it or not

lolz

 :P  :rotfl:  :rotfl:

@Jaswinder:
You can check online the status of your application.

ahh...  when  is the last date , .... i havent done anything!  :rotfl:

Coooeey!
U guys are really inspiring! :D
Do u guys ever feel tensed?? :huh:

[quote="nuclear_alchemist"][quote="accuracy"]can computer science students take the KVPY?[/quote]
Obviously. MOST of the scholars will be CS ppl. Bio/Hindi/Humanity students are logically idiots in general. Obviously, there are a few oddballs like that Rituraj <_<[/quote]

Thats very chavunistic....I know some smart kids who did take bio.

[hide]but only some... :D[/hide]

[quote="preetham"]these guys shud wrk on making the
KVPY well known like the INO's

many talented ppl dont  know abt kvpy and miss out on these scholarships[/quote]
I got 2 extra brochures from KVPY to give  to my school a couple of days ago :P
[/quote]

Thats very chavunistic....I know some smart kids who did take bio.

[hide]but only some... :D[/hide][/quote]
I said there are exceptions. 2 out my life's 3 loves have been bio girls :P And intelligence is always my foremost criteria. And I have many other friends too who are faaaar more intelligent than the usual maths students. I know what you mean.

[quote="nuclear_alchemist"][quote="accuracy"]can computer science students take the KVPY?[/quote]
Obviously. MOST of the scholars will be CS ppl. Bio/Hindi/Humanity students are logically idiots in general. Obviously, there are a few oddballs like that Rituraj <_<[/quote]

That's complete nonsense. Where do you think the world got insulin, the artificial heart, dialysis, open heart surgery et all if it weren't for your 'idiotic' biology people? Came as outputs of computer programs, did they?

You should take those blinkers off, they're limiting your vision as well as giving you a false sense of security and self-esteem.

hey guys have u got ur admit cards

i haven't yet mine

plz tell if u have received

thankx..

Surprised to see my name in the list..............
I didnt performed well in the examination but got selected

First of all, congratulations to all who have been selected. Well, those who didn't, don't give up and it's nothing but a silly exam after all where the strengths aren't measured accurately.

BTW when are the interviews held? And what are the centres?

thankx franklampard

don't be disheartened u will surely make it next year

congrats to all those selected

are 518 guys selected for interviews every year or is thsi yr different??

[quote]~500 every year increasing marginally
yeah last year it was 516,this year 518 :rotfl:  :rotfl: [/quote]

Lol , Im not disheartened guys. All good. Ive learnt not to think of an exam once its over ( INAO,INMO,INJSO) helped me last year  :P Like , I waited 2 months wasting time only to learn I dint make it in either of them.

To everyone selected , Best Of luck

3 from M.Prakash selected this year compared to last years 8 and two years back's 15  :rotfl:

Gotta work harder. :ninja:

Congratulations to all selected!!! Special one to Akashnil!!!- You are having an awesome year.
Aditya(franklampard) - Better luck next time..Wish you all the best for INMO 2010.

any1 wants to give final tips regarding NSEA and NSEJS ..my 1st time you see...n btw i find it diff to decide which 1 to concentrate on!!

BTW does any1 out here give NSO
P.S: How did u like today's ppr?

Congrats to all selected! :D
@sahil
Ya I used to sit for it till the previous year.......but I got frustrated with it as my ranks in it were never very decent.But I think I shud have appeared this year.
I saw the paper.... math questions were good.The 2nd 1.....did u tick option D? :huh:

oh!ya i did 2nd option..i think maths Q's were lame...and time consuming..only a mad amn can do 50 Q's in 60 mins..woa bhi good Q's...althou i left (thayt is did tukkabazi)in 4-5 maths Q's,i did this in nly 1-2 SC ones..resn being SC ones dont REQ much time...maths ones do..ppl tell me maths was VERY EZ..guess i shud've done'em....btw this one..didnt have mental ability ..it had maths..these guys lie..so i guess cut offs will be quite low this time

How cum u din tick option D (ques 2 math)Provided i've done no silly mistake...dat's the ans i got :huh:

dont u reed b4 typing??i said i ticked option D

[quote]oh!ya i did 2nd option[/quote]
So i thot u marked B :P


Is it possible to find $2005$ different positive square numbers such that their sum is also a square number ?

Let $m = 1^2 + 2^2 + ... + 2001^2 + 2002^2 + 2003^2 + 2005^2$
$m$ is odd, so $m = 2n+1$ for some integer $n$.
$m + n^2 = (n+1)^2$, and noting that $n > 2005$, the left side is the sum of $2005$ different positive square numbers.

I though about this about a half of hour and found that the following is also true:

There exist $0<a_1<a_2<\cdots<a_{2005}$ so that $\sum_{j=1}^ka_j^2$ are all squares for all $k=1,2,\ldots,2005$.  :)


If the ecuation $X^n-aX^{n-1}+1=0$, with $n\in N$, n>3 and a>2 has the real root $\alpha\in (0, 1)$, prove that any complex root that is not real $z$ has $|z|<\alpha$ (it is enough to prove that is <1)

Hello.

Let z = x.
$z^{n} - az^{n-1} + 1 = 0$

\[ z^n = az^{n-1} - 1 \]

\[ |z^n| = |az^{n-1} - 1| < |az^{n-1}| < |z^{n-1}|  \]

Next, note that for positive b, $b^{n} < b^{n-1}$ iff $0 < b<1$ for all natural n. Thus, conclude that $|z|^n<1, \ and \ that \ |z| < 1$.

Why did you say $|az^{n-1} - 1|<|az^{n-1}|$ ? Also, the inequality $|az^{n-1}| < |z^{n-1}|$ is not true because a>2. But if you tell me why $|az^{n-1} - 1|<|az^{n-1}|$, i will be able to solve the problem.

[quote="Frozen"]Why did you say $|az^{n-1} - 1|<|az^{n-1}|$ ?[/quote]
Using the triangle inequality maybe... :roll:

But the triangle inequality says $|az^n-1|\leq a|z^n|+1$

I apologize for the oversight. Please disregard my original post.

Apply Rouche's theorem, noting that $|az^{n-1}|>|z^n+1|$ for all $z\in(0,1)$.  This implies that $z^n-az^{n-1}+1$ has precisely $n-1$ roots inside the unit circle.  Putting $x=1$ and $x=a$ shows that there is a real root $t>1$.  This proves that the remaining roots have modulus less than $1$.

Can this problem be solved without complicated stuff like Rouche?

Here is my solution (inspired from Valentin's book). It is in romanian.


group theory
abstract algebra
superior algebra
superior algebra solved

if all subgroups of $G$ are normal, is $G$ necessarry abelian ?

No, take e.g. the quaternionic group $Q_8$.


calculus
integration
limit
real analysis
function
real analysis solved

Compute the limit:

$\displaystyle \lim_{n \to \infty} (2n+1) \int_{0}^{1} x^{n}e^{x} dx.$

Let $I_n = \int_0^1 x^n e^x dx$ and it is easy to find out that $I_{n+1} = e-(n+1)I_n$.

It is not difficult to prove that $I_n\to 0$ (write $\int_0^1$ as $\int_0^{1-\delta} + \int_{1-\delta}^1$; or apply Lebesgue's Convergence Theorem)

Hence $e-nI_{n-1}\to 0$, which implies $nI_n \to e$.

Therefore $(2n+1)I_n = 2nI_n + I_n \to 2e + 0 = 2e$.  :)

We might as well apply Moubinool's approach to the original problem:

$\lim_{n\to\infty}\frac{2n+1}n\int_0^1y^{1/n}e^{y^{1/n}}dy.$

Since $y^{1/n}\to1$ for all $y>0$, the integrand tends pointwise to $e.$ We could justify the interchange of limit and integral by the Dominated Convergence Theorem, or by various non-Lebesgue means. For instance: for any $\delta>0$, the sequence of functions converges uniformly on $[\delta,1],$ and the size of the integral itself, as well as the difference between the integrals, on $[0,\delta]$ is less than $e\delta$ and can be made as small as we please.

So the answer is $2e.$

[quote="Mitzah"]Can you please post the first proof in which you show that $I_{n}=0$ (not the one with Lebesgue)?

thank you very much![/quote]

Notice that $x^n e^x$ is increasing and bounded (the upper bound is $e$) on $[0,1]$.
For any given $\epsilon$, we can find such a $\delta>0$ such that $e \delta < \epsilon/2$.

$\int_0^{1-\delta} x^ne^x dx \leq \int_0^{1-\delta} (1-\delta)^n e^{1-\delta} dx = (1-\delta)^{n+1} e^{1-\delta}$
Obviously it tends to 0 as $n\to\infty$, hence for $n$ large enough, it is smaller than $\epsilon / 2$.

And $\int_{1-\delta}^1 x^ne^x dx \leq e \delta<\epsilon/2$.

Hence $\int_0^1 x^ne^xdx < \epsilon/2 + \epsilon/2 = \epsilon$ for sufficient large $n$.


calculus
integration
calculus computations

The additive property for single variable integral

$ \int_b^a {f(x)\,dx}  \equal{} \int_b^c {f(x)\,dx}  \plus{} \int_c^a {f(x)\,dx}$

Is there a similar property for double integrals?

Of course- the sum of the integrals over two disjoint regions is the integral over the union. This is not going to look as simple in the limits.

Yes: if you cut the region of integration into pieces, the integral is equal to the sum of integrals over each piece.
Sometimes you pretty much have to do this, because the given region of integration is not suitable for converting the double integral into an iterated integral. For example: $ D\equal{}\{(x,y)\colon |x|\plus{}|y|\le 1\}$ is a square turned by $ 45$ degrees. One can cut it by the vertical line through the center to get
\[ \iint_D f(x,y)\,dA\equal{}\int_{\minus{}1}^0\int_x^{\minus{}x}f(x,y)\,dy\,dx\plus{}\int_0^1\int_{\minus{}x}^x f(x,y)\,dy\,dx\]

Thanks for your answers.

[quote="mlok"]
\[ \iint_D f(x,y)\,dA \equal{} \int_{ \minus{} 1}^0\int_x^{ \minus{} x}f(x,y)\,dy\,dx \plus{} \int_0^1\int_{ \minus{} x}^x f(x,y)\,dy\,dx
\]
[/quote]
Then how would be $ \int {\int {f(x,y)\,dy} \,dx}$?

That's a meaningless notation. Double integrals must be done over a region.

Let $ D_1$ and $ D_2$ be two regions (defined here as non-empty bounded subsets of $ \mathbb{R}^n$) and $ D_1\cap D_2$ has null Lebesque measure then $ \iint_{D_1\cup D_2} f \equal{} \iint_{D_1}f\plus{}\iint_{D_2}f$ where $ f$ is integrable.


geometry
incenter
angle bisector
perpendicular bisector

In triangle ABC, I is the incentre. Let the incircle touch the sides AB and AC at Z and Y respectively. If AI meets the incircle in P, show that P is the incentre of triangle AZY

[hide]

cyclic quad

alternate segment theorem

[/hide]

[hide]$AP$ is the angle bisector of $\angle ZAY$.

So it's enough to prove that $YP$ is the angle bisector of $\angle AYZ$

$P,Z,Y$ belong to the incircle of $\triangle ABC$.

$\angle PYZ = \angle PYA \Leftrightarrow$

$arc\ PZ = arc\ PY \Leftrightarrow$

$PZ=PY$, which is true, because $AP$ is the perpendicular bisector of the segment $ZY$[/hide]


Can someone justify if this is the proof to why you cannot divide by 0?

I believe it's somewhere along the lines of the fact that if you divide by a number less than zero, the answer you get is larger than the original number.

For example, if you divided 3 by .00000000000000000000009, it would be an extremely large number. Since 0 is actually 0.00000000 and etc, the answer would be an infinately large number.

You are on the right track. However, I think you meant to say that when you divide by a number less than [i]one[/i], you get a number larger than what you began with (assuming you started with a positive number.

For example, 10/ .5 = 20

However, for negative numbers, it is reverse.

-10 / .5 = -20.

Dividing by a number less than one takes you farther away from the center of the number line.

Also think of it this way, dividing by a number less than one is like multiplying a number by something greater than one. This is because the reciprocal of a number less than one is greater than one. This is all concerning positive numbers, though.

A number divided by zero doesn't work ever.

If 50 / 0 = x

That would mean 0x = 50. And that is never true for any x.
I believe the answer for anything that is divided by zero is "undefined", not infinity, but I am not 100% sure on that.

Division by 0 can be undefined or plus or minus infinity depending on what's more convenient. This is the same as the fact that parallel lines can intersect in 0 or 2 places, depending on what is more convenient. Suppose that:
1/0=b
then 0b=1
recall that 0 times a number is 0. Some people say that infinity is large enough that the 0 doesn't affect it except to make it finite.

For all your purposes, dividing by 0 is undefined.

They should make a number $ g $ for 1/0. Like  $ \sqrt{-1} = i $

I agree with biffandoc.
sqrt(-1)=i because i is an [b]i[/b]maginary number
1/0 is a [b]g[/b]rand number
let's call 1/0 g.

you need to set up an equation. You want to prove that x=y/0.

Lets say y is six. x=6/0.

multiply both sides by zero. 0x=6

anything times zero is zero. 0=6

0=6 is not possible!!!

There is no G because its useless once you define it.
but making  :sqrt: -1 equal to i we open up many things we could not do before

If you put g on your math test for the problem 1/0, then your math teacher would probably not mark it right,

i don't know if trigonometry is too advanced for middle school, because it is 9th grade math here.
anyways, you can address the question by looking at tan(90degrees) or tan(.5pi radians)
tan, tangent, means the ratio of the two legs on a right triangle.
in this situation, one of the legs is equal to the hypotneuse and the other is zero, so tan(90degrees) is a/0.
but, look at this triangle. it is not a triangle!!!

[quote="NoSoupForYou"]If $g=\frac{1}{0}$, then $g^n=g$.

A new branch of math: G THEORY!![/quote]YAY!!

Wow, I didn't think peopel would find this so interesting! I'll post more topics


Can anyone help me with the following please? Thanks

If A is positive definite, show that each diagonal entry is positive.

I know that detA >0 and the fact that det((r)A) >0 for each r = 1, 2, ....n
but how should i show it's true?

Let $A$ be positive definite and $\vec{x}_{i}$ be a vector whose only nonzero coordinate is $\left[\vec{x}\right]_{i}=1$. Then $0< \vec{x}^{T}\cdot A\cdot\vec{x}= \left[A\right]_{i,i}$, so for any $i$ the $i$-th diagonal element of $A$ is positive.


function
number theory
greatest common divisor
SFFT
special factorizations

Find all integer solutions to:

$x+y=xy$

Theres two quick ways to solve it.

By symetry : $x = y \Rightarrow x^2 = 2x \Rightarrow (x, y) = (0, 0) \bf or (2, 2)$

$y = \frac{x}{x-1}, \quad \because gcd(x,x-1) = 1, \Rightarrow x - 1 = -1 \, or \, 1$

A few interesting ones there. mine was the same as your first one.... from the origianal equation, y divides x, and x divides y, so x=y. Proceed from here!

Since that was too easy, if

$x+y+z$  is a factor  $x^2+y^2+z^2+kxyz$  find k.

[quote="ankur87"][quote="svrsk"]$y = \frac{x}{x-1}, \quad \because gcd(x,x-1) = 1, \Rightarrow x - 1 = 0 \, or \, 1$[/quote]

In this equation, $x-1$ can't be equal to zero, since that puts a zero in the denominator of  $y = \frac{x}{x-1}|$[/quote]

Sorry. It must be $x - 1 = -1 \, or 1$  :P

[quote="seamusoboyle"]$x+y+z$  is a factor  $x^2+y^2+z^2+kxyz$  find k.[/quote]
[hide]Since $x + y + z$ is a factor of $x^2 + y^2 + z^2 + kxyz$, another factor must be $\frac {x^2 + y^2 + z^2 + kxyz}{x + y + z} = (x + y + z)(x + y + z - 4 + \frac {kxyz - 4z - 2xy}{x + y + z})$. So $( x + y + z - 4 + \frac {kxyz - 4z - 2xy}{x + y + z} = \frac {x^2 + y^2 + z^2}{x + y + z} \rightarrow {x^2 + y^2 + z^2 + 2xz + 2yz - 4x - 8z + kxyz} = x^2 + y^2 + z^2$. $x^2 + y^2 + z^2$ kill off leaving $kxyz = -2xz - 2yz - 4x - 8z$ so $k = \frac {-2(xz + yz + 2x + 4z)}{xyz}$.[/hide]
I think.

I figured that:

[hide]
you can square x+y+z to get x^2 + y^2 + z^2 + 2(xy + yz+ xz)

Therefore, you can set kxyz equal to 2(xy + yz+ xz), and so k will equal 2(xy + yz+ xz)/xyz.

[/hide]

[quote="RC-7th"][quote="svrsk"]
$x+y = xy \Rightarrow (x-1)(y-1) = 1$[/quote]

An application of Simon's Favorite Factoring Trick![/quote]

Who is Simon. ? Can you tell me. What is Simon's Favorite Factoring Trick! ;)

ComplexZeta is Simon.

$xy + x +y + 1 = (x+1)(y+1)$ is his favorite factoring trick, from what I've read...

$xy$ means $x*y$  ;)
you can't just read the two together :p

oh how stupid of me :D
i thought all x or y should be the same and the only thing you will do is add.
i thought it dont need to be multipied! (im talking about the sum)

[quote="cindy28"]i think the answer is:

x + y = xy
0 + 1 = 01
0 + 0 = 00

but this is just a guess :D[/quote]
well, if you look at it like this, all integers are possible  :P

[quote="seamusoboyle"]For the k, sorry for being obscure, find all integer values for k, for all x,y,z[/quote]

Oh. !!! very hard problem. I can work out for some case. :P

By Remainder Theorem : Substitute $z = -(x+y)$ in $x^2 + y^2 + z^2 + kxyz$
We will get : $x^2 + y^2 + (x+y)^2 - kxy(x+y) = 0 \Rightarrow k = \frac{2(x^2 + y^2 + xy)}{xy(x+y)}$

if $x=y = 1 \Rightarrow k = 3$

[quote="seamusoboyle"]Not sure what you did, k should turn out to be $-3$[/quote]

from your problem : if $x = y = 1, k = 3$ then $x^2 + y^2 + z^2 + kxyz = 1 + 1 + z^2 + 3z = (z + 1)(z + 2)$
and $x+y+z = 1 + 1 + z = z + 2$ so $(x+y+z)|(x^2 + y^2 + z^2 + kxyz $)

But if $x=y=1, k = -3$ then $x^2 + y^2 + z^2 + kxyz = (z - 1)(z - 2)$
but $x+y+z = 2 + z$ is not factor of $(z-1)(z-2)$ so $k \not= -3$

But if we change problem to $x^3 + y^3 + z^3 + kxyz$ in this case $k = -3$ since $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$  :blush:

[quote="djimenez"][quote="svrsk"]$ gcd(x,x-1) = 1, \Rightarrow x - 1 = -1 \, or \, 1$[/quote]

Just in case, this is totally false!, $\gcd(x,x-1) = 1$ for any integer $x\neq0,1$ (because in tose cases the function gcd is undefined).[/quote]

$gcd(a,b)=gcd(-a,b)=gcd(a,-b)=gcd(-a,-b) =$  positive interger number

$gcd(0,-1) = gcd(0, 1) = 1$  :blush: