It will take time to tell.  You have to give him at least a year before you can even think aboute valuating his progress.  I certainly hope he lives up to the hype.

i think he will make a good president. i am excited...  :)

[quote="nick42"]
so what do you guys think? will he live up to all the hype?[/quote]

No one could live up to the hype, and he better hope people remember that when we're still in Iraq in 18 months, and we're still having tough economic times at this time next year.  [i](Edited to note that I don't say this to bash Obama -- I'm merely noting that there's nothing he can do to change the short-term forecast of our involvement in Iraq or the economy.)[/i]

I think a big problem these days is that people vote for certain people because they are charismatic and have some kind of connection with voters.  According to a poll by Rasmussen (it was mentioned on Fox News on Tuesday so I'm assuming it was Rasmussen), more people agreed with McCain's policies, and more people see McCain as the better leader. But why did Obama win? He's obviously more smooth-talking and charismatic than McCain. Maybe there are other reasons, but that's the one that jumps out at me. I'm sure many of you may have voted for Obama because you agree with his proposed policies and think that he will follow through on them and be an effective leader. Good, that's the reason you should be voting for him.

Politicians make people believe that all their promises are going to magically happen. I have no actual evidence for this, but I see it is as possible that once Obama gets all the top-secret CIA briefings about foreign policy matters, he may have to change some of his positions. I reiterate, this is merely a guess and something I see as possible.

[quote="CatalystOfNostalgia"]But why did Obama win?[/quote]

I have a one-factor model to describe this: Because Obama is much farther from George Bush than McCain is.  After the last 8 years, that's tough logic to defeat.

Also, it's pretty commonly known that Fox is a conservative news channel, so you should expect some bias.

[quote="nick42"]he just picked Rahm Emanuel for his chief of staff.[/quote]
Fans of the television show [i]The West Wing[/i] note that Matt Santos (Obama) defeated Arnie Vinick (McCain). But Obama picked Josh Lyman (Emanuel) for Chief of Staff, not Leo McGarry (Tom Daschle). Of course in the fiction McGarry, who had been COS for the outgoing President Bartlet, was on the Santos ticket as the VP nominee, but died before he could take office. (Because the actor who played him died.)

[quote="CatalystOfNostalgia"]I think a big problem these days is that people vote for certain people because they are charismatic and have some kind of connection with voters.[/quote]

It's not just Barack Obama's charisma -- it's his attitude. He's calm, he's cool, and he's unflappable. These traits are not just attractive, they are important for a president to exhibit in his (so far) day to day dealings and his responses to emergent situations.

It's the stuff like citing Reagan as a model of change that really makes me like Obama. Call me naïve, but it does seem like he will be less partisan than the hyperpartisan Bush administration.

When we look at it, really, it would have been very difficult for McCain to win. Three main factors favored Obama:

1. President Bush, who is viewed by the American people as in general a pretty bad president. This made it difficult for any Republicans.

2. An incredibly unpopular war which was denounced from the beginning by one candidate and continued to be steadfastly endorsed by the other.

3. An economic crisis - McCain had an advantage on foreign policy, but the crisis distracted the news from anything overseas. Obama's demeanor and consistent message appealed more to voters than McCain's apparent attempt to throw everything at the wall and see what stuck. McCain's comment about the fundamentals of the economy was just stupid. Regardless of truth, it was a dumb thing to say, and it cost him.

Moreover, Obama ran a much, much better campaign than McCain. McCain was forced to spend money on making some attempt to maintain advertising parity, so his ground game suffered. Having worked on the Obama campaign, not even a swing state, it was really well run. It was very inclusive and even meritocratic, as the best volunteers were put to work organizing others. His message was always steady, but McCain was flitting from attack to attack, finally finding some semblance of a message with the "socialism" deal; however, this was not extremely credible and only served to rev up the base. The Palin pick was another move to gain support from the base, but it ended up being disastrous. Then again, we don't know what would have happened had McCain picked Lieberman, Ridge, or even just Pawlenty. Palin was clearly not ready for national politics. While she seems a pretty savvy politician, she did not have the experience to be an appealing candidate, let alone an effective vice president. It showed.

For a partial answer to the question of what Obama will do right away, there's [url=http://www.washingtonpost.com/wp-dyn/content/article/2008/11/08/AR2008110801856.html?hpid=topnews]this[/url]:
[quote="Washington Post"]Transition advisers to President-elect Barack Obama have compiled a list of about 200 Bush administration actions and executive orders that could be swiftly undone to reverse White House policies on climate change, stem cell research, reproductive rights and other issues, according to congressional Democrats, campaign aides and experts working with the transition team.[/quote]

While some of the hype is ridiculous or even contradictory, there's a whole lot of direct executive action possible, and the Obama team has already been planning for months. After eight years of Bush, this kind of focused competence might be a shock.

let me ask u this, though:
Do you believe that the government should take money from hardworking people, and give it away to those who do not work? Ok, maybe there are some people who cannot financially support their families, but still. The majority of those who do not work choose to not work. I don't know about others, but I think it is vastly unfair to those who work hard to get a steady income; they should be able to enjoy the fruits of their labor. Instead, people who are lethargic and choose not to work will receive the former's income.  And these people range from criminals, gangsters, and illegals, most likely. And Obama wants to do this.  This will trigger a mass effect called the 'Work Massacre', in which work will be massacred.  No one will think that there's a point in working, because the government will come and take away a decent chunk of their income. So thus, if no one works, there's no income, and no money the government can take, and no money the government can give, and then the US collapses.  Great plan :| .
All I can do is pray to God that this stuff will not happen.

[quote="bbgun34"]And these people range from criminals, gangsters, and illegals, most likely. And Obama wants to do this. [/quote]

Haha "illegals"? I assume you mean illegal immigrants? Trust me, they work A LOT harder than most citizens-- they are the people who work 14 hours a day yet they get paid probably four dollars per hour. Talk about work being massacred.

[quote="bbgun34"]let me ask u this, though:
Do you believe that the government should take money from hardworking people, and give it away to those who do not work? Ok, maybe there are some people who cannot financially support their families, but still. The majority of those who do not work choose to not work. I don't know about others, but I think it is vastly unfair to those who work hard to get a steady income; they should be able to enjoy the fruits of their labor.[/quote] I don't get why the right is getting so worked up over this. Obama is planning on raising the top level's taxes 3%. That's it. He isn't raising anyone else's taxes.

If you think 39% is that high, then look at this:

http://ataxingmatter.blogs.com/.a/6a00d8341cf2a753ef01127967811f28a4-800wi

Oh, yeah, taxes. How much did I pay for 2008? A whole lot less than for 2007. This is not a good thing; I'd rather have the income to pay taxes on.

Well I'd assume that part of the reason you (and everybody) has less money this year was because of those mentioned tax cuts.

[quote="bbgun34"]The majority of those who do not work choose to not work.[/quote]
I highly doubt that. Do you have any evidence¿

[quote="bbgun34"]I don't know about others, but I think it is vastly unfair to those who work hard to get a steady income; they should be able to enjoy the fruits of their labor. Instead, people who are lethargic and choose not to work will receive the former's income.[/quote]
You know that, as their work doesn't really pay back for society, you can stop paying universeity professors of mathematics, philosophy, ...  :wink:

[quote="bbgun34"]And these people range from criminals, gangsters, and illegals, most likely. And Obama wants to do this.[/quote]
_Any_ evidence¿

[quote="bbgun34"]No one will think that there's a point in working, because the government will come and take away a decent chunk of their income. So thus, if no one works, there's no income, and no money the government can take, and no money the government can give, and then the US collapses.  Great plan :| .[/quote]
We have that since about 60 years and I may have missed germany's total collapse... At least it's not us or this reason causing the current crisis  :wink:

ZetaX -- evidence?  How could he go around repeating only minimally intelligible talking points if he were to let himself be encumbered by evidence?  ;)

JBL-you say his points are "minimally intelligible."  Don't you think you're giving him too much credit? :wink:  These sound like things Rush Limbaugh would say.

[quote="evilhamster"]I agree with jmerry. Obama is probably one of the only people in the world capable of reversing eight years of Bush policies. However, this is not because he's a better person than McCain, or even a better candidate for office. Right now, he's better than McCain, because he has the momentum of the country on his side, as well as an increasingly liberal congress to work with. Obama is liberal, and change in policy is what this country wants (and needs).The Christian right has come out bad in this election, and it will be hard for them to keep Obama from implementing his plans.
That said, his plans had better be beneficial. Elections are coming up in two years, and that's pretty much the deadline for Obama. If he and Congress can't seriously overhaul the nation's reputation in that time, he's bound to lose his power.
But I believe Obama is up to the challenge.[/quote]

All of these are good observations. Obama is indeed capable of reversing the worst of Bush's policies, whether you believe it's the Iraq Occupation, Guantanamo, large deficit spending, etc. However, the fundamental doctrines under which our government is run today will not change.

The United States will still attempt to "manage" world affairs, even if we draw down troops in Iraq. We will still have troops in 2/3 of the world's countries; we will still be committed to protect other industrialized countries; our defense budget will continue to increase, despite the fact that the U.S. spends more on defense than the rest of the world combined.
Has our basic foreign policy been beneficial for the security of Americans? This must be reevaluated. And I believe the answer is no. Many of our actions around the world (whether it's the 1953 CIA-assisted Coup that ousted Mossadegh in Iran, our military presence in Saudi Arabia, our unconditional support of Israel) have actually jeopardized our security.

The United States government will continue to live beyond its means. Obama's 2009 budget will likely have a \$1.84 trillion deficit. He talks about cutting government waste but instead quadruples the deficit. Granted, many of these funds go toward socially laudable causes. But do we really want the government to manage more of the economy?

Indeed, the country may desire a swing to the left. But will this significantly improve our foreign policy? I don't think so. Also, I believe that his economic policies can actually turn out worse than those of Bush.

The Christian Right has largely been outvoted in this election, and it will indeed be difficult for them to maintain their social policy. Seeking to restrict is seeking to exclude. If they are to gain more credibility, I believe the Christian Right must adopt the traditional (George Washington's Farewell Speech) positions on foreign and economic policy.

[quote="camelNotation"]The Christian Right ... the Christian Right ...[/quote]
Do you think there is a difference between the Right and the Christian Right???
I do.

Yes, the Christian Right is not the Right. I pointed out the Christian Right as a response to evilhamster's statements.

Now that I think it through, I believe evilhamster meant to say the Right in general.


Solve in positive integers the equation \[ n = \varphi(n) + 420 ,  \] where $\varphi(n)$ is the number of positive integers less than $n$ having no common prime factors with $n$.

[quote="ZetaX"]It looks very similar to http://www.mathlinks.ro/Forum/viewtopic.php?t=71348 ...[/quote]Yeah, only this one is with 420 instead of 402 due to a spelling mistake. Still solvable :P so thread was reopened.

This is certainly not the most elegant of solutions.

Also, I am uncertain if the problem with 402 is still open.  There are many ideas on the link mentioned above.  Whether or not a complete proof exists, I am unsure?

[quote="Valentin Vornicu"]Solve in positive integers the equation $ n \equal{} \varphi(n) \plus{} 420$, where $ \varphi(\cdot)$ is the number of positive integers less than $ n$ having no common prime factors with $ n$.[/quote]

I have posted my solution (without any computer use) here:

[url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem 13.A equation involving EulerPhi-part 2[/url].

I have also posted the solution in the case when $ 420$ is replaced with $ 402$ as [i]Problem 12.A equation involving EulerPhi-part 1[/i] at the same link.

Does someone know if this study of boundedness of $ |\{n \in \mathbb{N}_0: n \minus{} \varphi(n) \equal{} k_0\}|$ with $ k_0 \in \mathbb{N}$ fixed is already done?
Cheers, Paolo  :)

If $ n\equal{}\phi(n)\plus{}2a$, then $ n\equal{}2m$ is even and $ a<m\le2a$, because $ \phi(n)$ odd only if $ n\equal{}1$ or $ n\equal{}2$. Therefore $ n\ge 2a\plus{}2$ and $ \phi(n)\le n/2\to n\le 4a$.
$ \phi(m)\equal{}2\mod 4$ only if $ n\equal{}4,p^k,2p^k$ for some prime $ p\equal{}3\mod 4$. If $ a>1$ $ n\equal{}4$ does not work. If $ n\equal{}p^k$, then $ p^k\not \equal{}\phi(n)\plus{}2a$, if $ n\equal{}2p^k$, then $ p^{k\minus{}1}(p\plus{}1)\equal{}2a$. In these case $ 420\equal{}2^2*3*5*7$ had 2 prime factors form $ 3\mod 4$, and $ n\equal{}2*3^2,2*7^2$ is not solution. Therefore $ 4|\phi(n) \to 4|n$.
 If $ 8|n, n>8$, then  $ 8|\phi(n)\to 8|420$. Therefore $ n\equal{}4*m$, $ m is odd$ and $ 105<m<210$.
If $ p^2|m$, then $ p\equal{}3$ or $ p\equal{}5$ or $ p\equal{}7$ and $ p^3\not |m\to (2p,m)\equal{}1$. We had $ 4p^2m\equal{}2p(p\minus{}1)\phi(m)\equal{}4*3*5*7$ or $ pm\equal{}\frac{p\minus{}1}{2}\plus{}\frac{3*5*7}{p}.$
If $ p\equal{}3$, then $ 3m\equal{}36$ not solution, $ p\equal{}5: 5m\equal{}23$ not solution, $ p\equal{}7: 7m\equal{}18$ - not solution.
Therefore $ n\equal{}4m$ and $ (420,m)\equal{}1, p^2\not m$. Therefore $ m\equal{}p$ or $ m\equal{}pq$.
Case $ m\equal{}pq$. In these case  $ pq\equal{}\frac{(p\minus{}1)(q\minus{}1)}{2}\plus{}105$ and $ pq<210$ only $ p\equal{}11,q\equal{}17$ - not solution.
Therefore $ m\equal{}p$ and $ 2p\equal{}p\minus{}1\plus{}210$ or $ p\equal{}209\equal{}11*19$ - not solution.


I just looked at my solutions online right now and realized the USAMTS ID I put in was not right.  What should I do?

Email usamts@usamts.org with the details.


AMC
AIME
USA(J)MO
USAMO
AMC 10
geometry
summer program

AMC 10 #25.

Typically, it seems #25 is about AIME #5 difficulty.

And AIME #12-15 is about USAMO #1 difficulty...

but AMC and AIME are different, adn AIME and USAMO are different...

note that AIME #1 is not necessarily the easiest one for everyone. And sometimes the amc10 #25 isnt the hardest one but generally the hardest one on the amc10 is much harder than the easiest one on the AIME, especially given the time restraints on the amc

Personally, just by nature of the test, the AIME #1 is harder, because in the AMC you have several choices, so if you cant work the problem out, you can use the assume each of the options are true in turn, and try to find more information until either everything fits, or something doesn't fit, if something diesn't fit, its not the answer, if everything fits, then its the answer

Yes, but AIME #1 is almost a typical school problem usually...

Heh...I couldn't solve #1 on the AIME this year, but I got #25 on the AMC 10...

i agree with georgiamathdude. it is hard to say one problem is easier than the other. for example, say i am really really really good @ counting (which is so untrue, i cannot do counting to save my life) and i get #15 on AIME using a very slick counting trick. Let us now say that i am really really really really bad @ geometry (true) and i miss #1 on AIME because it was geometry. to me #15 would be easier because i knew how to do it cuz i am good at counting, than #1 which i didn't know how to do because i am not good at geometry.

but i guess #25 on AMC 10 is more difficult (for me) than AIME #1. Difficulty of a problem is a very personal thing.

Also, you can't just say #1 is easier than #2.  But you can usually say that #1 is easier than #20.

btw anyone else interested in PROMYS 2009?


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Prove that if $f$ is a continuous real-valued function on the interval $[a,b]$ such that $f(x) \ge 0$ for all $x\in[a,b]$ and $f(x)>0$ for some $x\in[a,b]$, then $\int^{b}_{a}f(x)\,dx > 0$.

Can anyone help me?

It's been posted, along with several harder variants.

The key to the proof: "stick a box under the graph."

Pick $\epsilon >0$ such that there exists $\delta > 0$ so y in $B(x;\delta$) implies $f(y)>\epsilon$ thus $\int^{b}_{a}f(x)\, dx>\int^{x+\delta}_{x-\delta}f(x)\, dx>\epsilon * \delta > 0$

mikesdaniel, can you please elaborate on your proof? Thanks for helping though.

Pedagogically speaking that is a pretty poor proof I think.  If $f(c)>0$, pick $\delta>0$ such that $|f(t)-f(c)|<f(c)/2$ for all $t$ such that $|t-c|<\delta$, e.g, for $t\in B(c,\delta)$ using mikesdaniel's notation.  The existence of such a $\delta$ follows from the continuity of $f$.

Then we have
\[\int^{b}_{a}f(t)dt \geq \int^{c+\delta}_{c-\delta}f(t)dt \geq \frac{f(c)}{2}\int^{c+\delta}_{c-\delta}dt=f(c)\delta \]
The last inequality is true since $f(c)>0$ and $|f(t)-f(c)|<f(c)/2$ implies $f(t)\geq f(c)/2$, and the interval $[c-\delta,c+\delta]$ has 'measure' $2\delta$.  Since $f(c)\delta>0$ we are done.


search
inequalities unsolved
inequalities

x,y,z are real numbers.prove that:

(xy + xz + zy)(1/(x+y)^2  + 1/(x+z)^2 + 1/(z+y)^2)>= 9/4

it's the well-kowned iran 96 's tst  :lol:  you can see here:

: http://www.mathlinks.ro/viewtopic.php?search_id=838963879&t=32470


Sequence
quadratics
relatively prime
IMO Shortlist
number theory

Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by \[x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots,\] has all of its terms relatively prime to $m$.

[i]Proposed by Jaroslaw Wroblewski, Poland[/i]

Let $u,\bar u$ be the roots of the equation $x^2-x-1=0$ (the characteristic equation of the sequence). Since every prime divisor of $m$ has $5$ as a quadratic residue, if we fix $p$, we can regard $u,\bar u$ as residues $\pmod p$, and carry out the computations in this manner.

The general term of the sequence is $x_n=\alpha u^n+\beta \bar u^n$ (where I changed the notations a bit and took $x_0=a,x_1=b$), and we must show that we can pick $\alpha,\beta\pmod p$ s.t. $\alpha u^n+\beta \bar u^n\ne 0,\ \forall n$ in $\mathbb Z_p$. We can do this by taking $au\equiv b\pmod p$, which translates (after doing the computations) to $\beta\equiv 0\pmod p$.

[quote]Since every prime divisor of m has 5 as a quadratic residue, if we fix p, we can regard $u,\bar u$ as residues $\pmod p$, and carry out the computations in this manner. [/quote]
I' m beginner in "mathematic english"  :blush: Can you explain me what does "quadratic residue" mean ?

jak sama nazwa wskazuje, jest to reszta kwadratowa

Can you give me a more clearly solution ,Grobber ?

Nice solution Grobber, it is also what I had!

It is a very nice solution!
One can make it a little bit more elementary.
If we take $ x = 2k + 1 + m $ then $ x^{2} - 2x - 4 \equiv
(2k+1)^{2} - 2(2k+1) - 4 = 4k^{2} + 4k + 1 - 4k - 2 - 4 = 4k^{2}-5 = m \equiv 0 \ (mod \ m) $.
Therefore because of $ x $ being even we can take $ y = x/2 $ and then
$ x^{2} - 2x - 4 = 4(y^{2} - y - 1) \equiv 0 \ (mod \ m) $, and because $ m $
is odd, we get that $ y^{2} - y - 1 \equiv 0 \ (mod \ m) $. Also we have
$ gcd(y,m)= 1 $, because $ 2y \equiv 2k+1 \ (mod \ m) $, so
$ 2(2k-1)y \equiv 4k^{2} - 1 \equiv 4 \ (mod \ m) $, and $ m $ is odd.
Therefore also $ y^{n} $ is co-prime to $ m $. Now if we take
$ x_{1} = y $, $ x_{2} = y^{2} $ and define $ x_{n} $ by recursion
$ x_{n+2} = x_{n+1} + x_{n} $, then of course $ x_{n+2} \equiv x_{n+1} + x_{n} \ (mod \ m) $,
but also because of $ y $ satisfying $ y^{2} - y - 1 \equiv 0 \ (mod \ m) $, we get that
$ y^{n+2} \equiv y^{n+1} + y^{n} \ (mod \ m) $, and by induction one can show that
$ x_{n} \equiv y^{n} \ (mod \ m) $ for every $ n \geqslant 0 $. Now,
we have seen that $ y^{n} $ is co-prime to $ m $, so $ x_{n} $ is also co-prime to $ n $.

Incidentally, this was also problem 6 of the 2nd TST of Taiwan 2005.

Suppose $k$ is not a multiple of 5, then let prime factorization $m=p_1^{a_1}p_2^{a_2}...p_r^{a_r}$.
Define the Fibonacci sequence to be $F_0=1, F_1=1, F_n=F_{n-1}+F_{n-2}$.
By Binet's formula,
$F_n = \frac{1}{\sqrt{5}} \left ( \left (\frac{1+\sqrt{5}}{2} \right )^{n+1} - \left ( \frac{1-\sqrt{5}}{2} \right )^{n+1} \right )$

Lemma 1: $x^2 = q^2(\mod p)$ has at two solutions $x$ modulo $p^a$ if $(p, q)=1$ and $p$ is odd prime.

Proof: It factors to $(x+q)(x-q)=0(\mod p)$. then $p$ divides either $x+q$ or $x-q$  the desired lemma follows.

Hence from Lemma 1 we know $\sqrt{5} = \pm 2k (mod p_i)$ for $i=1, 2, ..., r$ since we know $p_i$ is not 2 or 5.

Lemma 2: $F_n \neq \left ( \frac{1+\sqrt{5}}{2} \right)F_{n-1} (\mod p_i)$.

Proof: Assume it's true, then  we know $\sqrt{5}=\pm 2k$ and it's relatively prime to $p_i$ so WLOG we let it be $\sqrt{5}=2k$ we substitute $F_n$ and $F_{n+1}$ and simplify we have

$(1-2k)^n (4k) = 0(\mod p_i)$. but if $p_i|1-2k$ then $p_i|m+(1+2k)(1-2k)$ => $p_i | 4$=> $p_i=2$ so contradiction. If $p_i | 4k$ then $p_i |5$ also contradiction.
The desired lemma follows(you could just make a simple argument with limits, but I'm not sure how rigorous that is)

Now let
$a=2k^2-k-3$
$a = -\frac{1+2k}{2} (\mod m)$  and so $a$ is relatively prime to $m$ since
$4k^2-5 - (2k+1)(2k-1) = -4$ and $(4, 4k^2-5)=1$

$b=1$ . Now $x_n=F_{n-1}a + F_{n})b=-F_{n-1} \phi + F_n(\mod p_i)$ and it's never 0 relatively prime to all $p_i$ so relatively prime to $m$ as desired.

If $k$ is a multiple of 5 it's not big deal, as 5 is the largest power of 5 that divides m so $m=5 p_1^{a_1} p_2^{a_2}...p_r^{a_r}$ so we can just say

$a=2k^2-k-3$
$a = 2(\mod 5)$ and $a = -\frac{1+2k}{2} (\mod \frac{m}{5})$

$b=1$, and we know $x_n$ is relatively prime to all $p_i$ and is relatively prime to 5 (modulo 5 $x_n$ sequence is 2, 1, 3, 4, 2, 1, 3, 4....). So $x_n$ and $m$ are relatively prime too.

QED

[quote="grobber"]The general term of the sequence is $ x_n \equal{} \alpha u^n \plus{} \beta \bar u^n$ (where I changed the notations a bit and took $ x_0 \equal{} a,x_1 \equal{} b$), and we must show that we can pick $ \alpha,\beta\pmod p$ s.t. $ \alpha u^n \plus{} \beta \bar u^n\ne 0,\ \forall n$ in $ \mathbb Z_p$. We can do this by taking $ au\equiv b\pmod p$, which translates (after doing the computations) to $ \beta\equiv 0\pmod p$.[/quote]

Can you explain in detail？ :maybe:

Here's another solution:

[hide="Solution"]
Define $|x|$ to be an integer $y$, $0 \leq y < m$ such that $x \equiv y \pmod{m}$.

Consider the sequence with $a=1$ and $b=\left|\frac{1+\sqrt{5}}{2}\right|$.

Note that $b^2=b+1$

Thus, our terms are $1,b,b^2,b^3,b^4....$; all we must show then is that $b$ is relatively prime to $m$.

Now, $\sqrt{5} \equiv \sqrt{4k^2} \equiv 2k \pmod{m}$.

Alas, $2k$ is even, so $\frac{1+\sqrt5}{2}\equiv \frac{1+2k}{2} \equiv \frac{1}{2}+k \pmod{m}$ is no integer!

However, $\frac{1}{2} \equiv \frac{4k^2-4}{2}\equiv 2k^2-2 \pmod{m}$, so

$b \equiv 2k^2+k-2 \pmod{m}$.

By the Euclidean Algorithm, $GCD(4k^2-5,2k^2+k-2)=GCD(2k+1,2k^2+k-2)=GCD(2k+1,-2)=1$.

Thus, $b$ and $m$ are coprime, so all terms of the sequence are relatively prime to $m$.[/hide]

[quote="abdurashidjon"]Is Leva 1980's solution is true? I could not translate to any of $a$ or $b$.
 Abdurashid[/quote]

$a=1$ and $b=y$.

[hide="Another possible approach"]By easy induction one has $x_n=f_nx_1+f_{n-1}x_0=f_na+f_{n-1}b$.
where $f_n$ is the n-th term of the Fibonacci sequence.

Modulo $4$ the residues of the sequence $<f_n>_{n\ge0}$ repeats at $1,1,2,3,1,0$.

Is it posible using this fact????[/hide]

Bye...

Sayantan...

[b]Claim.[/b] There is a natural number $b$ such that $b^2=b+1 \pmod{m}$.
[b]Proof of Claim.[/b] As $m$ is an odd number, it suffices to find a $b$ that $(2b-1)^2 \equiv 5 \pmod{m}$
Set $m=\prod_{i=1}^k p^{e_i}_i$. For each $p_i$, as $(2k)^2 \equiv 5 \pmod{p_i}$, we have $\left( \frac{5}{p} \right)=1$.
Now there exists an $x$ such that $x^2 \equiv 5 \pmod{p_i}$ for each $i$.
By Hensel's lemma on $f(x)=x^2-5$, we can see that there exists an $x$ that $x^2 \equiv 5 \pmod{p^{e_i}_i}$.
By CRT, we can ensure that there exists an $x$ that $x^2 \equiv 5 \pmod{m}$, and we conclude.

Now set $x_0=1$ and $x_1=b$. Then $x_n \equiv b^n \pmod{m}$, so $(x_n,m)=1$ as desired.

umm... isn't this just $a=2,b=2k+1$?

Take any prime $p$ that divides $4k^2-5$
then $(2k)^2 \equiv 5 (mod p)$

using characteristic polynomials, we get $x_n \equiv 2*(\frac{1+2k}{2})^n (mod p)$

since $p$ is odd and obviously $2k$ is not $-1(modp)$ so $x_n$ coprime to $p$ so it is also coprime to $4k^2-5$ for all n.

Solution:
 We can take $a=1, b=2k^2+k-2$.
Lemma: If we consider the sequence $x_1=1, x_2=s, x_n=x_{n-1}+x_{n-2}$ then all terms of $x_n$ are co-prime to $s^2-s-1$.
Proof: Any term of $x_n$ is in the form $sF_r+F_{r-1}$.
Now, note that $F_r^2(s^2-s-1)-(sF_r+F_{r-1})(sF_r-F_{r+1})=F_{r-1}F_{r+1}-F_r^2$ which is always of the form $\pm 1$ by Cassini's identity, thus any $x_n$ is coprime to $s^2-s-1$.
If we put $s=2k^2+k-2$ note that $(2k^2+k-2)(2k^2+k-3)-1=4k^4+4k^3-9k^2-5k^2+5=(4k^2-5)(k^2+k-1)$ thus for $a=1, b=2k^2+k-2$ the terms are always coprime to $4k^2-5$.
@3below(math_pi_rate): The lemma is not true for $p=3$

Nice problem! Here's a different solution (Hopefully correct!): Let $F_n$ denote the Fibonacci sequence with $F_{-1}=1$ and $F_0=0$. It's easy to see (using induction) that $$x_n=aF_{n-1}+bF_n \quad \forall n \in \mathbb{N} \cup \{0\}$$
The crucial lemma is as follows:-

[b]LEMMA[/b] For all primes $p$ such that $5$ is a quadratic residue modulo $p$, there exists some $C \in \mathbb{Z}/ p\mathbb{Z}$ such that $$F_n \not \equiv CF_{n-1} \pmod{p}$$
for any $n \in \mathbb{N} \cup \{0\}$.

[u][i]Proof of Claim[/i][/u] If $p=5$, then one can easily check that $C=3$ works (since in this case, one finds that $F_n$ is periodic modulo $5$ with period $4$, which is not true). So assume $p>5$, and suppose the Lemma is not true. Then, for all $i \in \mathbb{Z}/ p\mathbb{Z}$, there is some index $n_i \geq 0$ such that $$F_{n_i} \equiv iF_{n_i-1} \pmod{p}$$
For all such $i$, let $n_i$ be the smallest such index. Since $\left(\frac{5}{p} \right)=1$, so there exists some $y$ such that $y^2 \equiv 5 \pmod{p}$. Then, by Binet's Formula, \begin{align*} F_n &=\frac{1}{\sqrt{5}} \left(\left(\frac{1+\sqrt{5}}{2} \right)^n-\left(\frac{1-\sqrt{5}}{2} \right)^n \right) \\ &\equiv \frac{1}{y} \left(\left(\frac{1+y}{2} \right)^n-\left(\frac{1-y}{2} \right)^n \right) \pmod{p} \\ &\equiv \frac{1}{y} \left(\left(\frac{1+y}{2} \right)^{n+p-1}-\left(\frac{1-y}{2} \right)^{n+p-1} \right) \pmod{p} \\ &\equiv F_{n+p-1} \pmod{p} \\ \end{align*}
which means that $F_n$ is periodic modulo $p$, with its period $T$ dividing $p-1$. In particular, this gives that $F_{n+T}=F_n$ for some $1 \leq T \leq p-1$. Since $n_0,n_1, \dots ,n_{p-1}$ are the smallest indices by definition, so due to periodicity after $p-1$ indices, we get $0 \leq n_i \leq p-2$. But then, by Pigeonhole Principle, two of these indices must coincide. Suppose $n_i=n_j$ with $i<j$. Then we get $$jF_{n_i-1}=jF_{n_j-1} \equiv F_{n_j} \equiv F_{n_i} \equiv iF_{n_i-1} \pmod{p} \Rightarrow (j-i)F_{n_i-1} \equiv 0 \pmod{p}$$
As $0<j-i<p$, so we have $$p \mid F_{n_i-1} \Rightarrow F_{n_i} \equiv iF_{n_i-1} \equiv 0 \pmod{p} \Rightarrow p \mid \gcd(F_{n_i},F_{n_i-1})$$
However, this contradicts the well known fact that $\gcd(F_r,F_{r-1})=1$ for all $r \in \mathbb{N} \cup \{0\}$. Thus, we have a contradiction, proving the Lemma. $\Box$
--------

Return to the problem at hand. Let $p_1,p_2, \dots, p_s$ be all distinct prime divisors of $m$. Then $5 \equiv (2k)^2 \pmod{p_i}$ gives that $5$ is a quadratic residue modulo $p_i$ for all $i \in [s]$. By our Lemma, there exist constants $C_1,C_2, \dots ,C_s$ such that $$F_n \not \equiv C_iF_{n-1} \pmod{p_i} \quad \forall n \in \mathbb{N} \cup \{0\} \text{ and } i \in [s]$$
Using CRT, there exists an integer $Z$ satisfying $$Z \equiv C_i \pmod{p_i} \quad \forall i \in [s]$$
Choose $b=1$ and $a>0$ with $a \equiv -Z \pmod{m}$. Then, for all $i \in [s]$, we get $$x_n=aF_{n-1}+bF_n \equiv F_n-ZF_{n-1} \not \equiv 0 \pmod{p_i} \Rightarrow \gcd(x_n,m)=1 \quad \forall n \geq 0 \text{ } \blacksquare$$

[quote=Al3jandro0000]@above your lemma is valid for any $p $, notice you're not using the fact $\left (\frac {5}{p}\right)=1$.[/quote]

I am using it in the proof of the Lemma while writing $\sqrt{5}$ as $y$. Note that if $5$ was not a quadratic residue, then I would have to work in $\mathbb{Z}[\sqrt{5}]$, and so Fermat's Little Theorem won't be applicable (Lagrange's Theorem gives a higher bound on the order, which isn't compatible with my PHP usage).

On the orther hand, there might be a possibility of the Lemma being true for all primes. I wonder if someone can prove/disprove that?

Tbh I don't think the Lemma is true always (In fact, one can easily check that it fails for $p=3$). However, it can be interesting to determine for what values of $p$ it works. Here are some comments PMed to me by Superguy (I haven't read it thoroughly, but it seems ok at first glance).

[quote=Superguy in First PM]Okay so far I have approached problem a bit.
First of all I have neglected the $F_{-1}$ part which won't change much of the proof
Assume ftsoc that there doesn't exist such a $C$
By considering the group $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ we can see that if $\left (\frac {5}{p}\right)=-1$ then $p$ is a prime in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$
Now using some tricks u can get that $a^{(p+1)}\equiv -1 \pmod{p}$ where $a$ is golden ratio.
Using this u can easily get
$F_{n+p+1}\equiv -F_{n}\pmod{p}$
We get that period is atmost $2p+2$
So considering the sets $(F_{1},F_{2},.....F_{p+1})$ and $(F_{p+2},F_{p+3}......F_{2p+2})$ and using the minimal index notation one can see that $2\leq n_{i}\leq p+1$
This means that there should be  $p$ i's given by $p$ $n_{i}$'s which implies each member of first set should give a separate $i$

For instance in case of $p=3$ clearly no such constant exists while in the case $p=17$ we clearly have a constant as $F_{9}\equiv F_{18}\equiv 0\pmod{17}$
Edit:See below for a bit more detailed explanation.[/quote]

[quote=Superguy in Second PM]Hmm a bit more
So clearly there should not be any index $0<i<p+1$  such that $p|F_{i}$ otherwise we can get our constant from above argument.
Now assuming for no such index $F_{i}$ is multiple of $p$.
Hence all $(\frac{F_{i}}{F_{i-1}})$ are uniquely defined modulo $p$
Now assuming that two of them are congruent then
$\frac{F_{i}}{F_{i-1}}\equiv \frac{F_{m}}{F_{m-1}}\pmod{p}$ for some $m>i$
Now manipulating terms we get
$(\frac{1+\sqrt{5}}{1-\sqrt{5}})^{m-j}\equiv 1\pmod{p}$.
Now clearly we can show that order divides $2(p+1)$ and hence if order is less than $(p+1)$ then we can clearly get $(m,j)$ and thus our desired constant however in case order is equal to $2(p+1)$ or $(p+1)$ we can't get our desired constant.[/quote]

Others can comment on this if they find something interesting :)

Well last part can be reduced to the fact that such a constant exists only if $F_{p+1}$ doesn't has a primitive prime divisor as $p$ which I don't think is possible to prove.
For instance after $p=3$ next such prime for which no such constant exists is $p=7$
followed by $p=23$

[color=#f00]Lemma.[/color] If $p$ is a prime such that $p|4k^2-5$ then we can pick $a,b$ such that for all $N\in\mathbb N$ we have $p\nmid x_n$.
[color=#f00]Proof.[/color]
[color=#0f0]CASE I:[/color] $p=5$. It suffices to take $a=1$ and $b=3$
[color=#0f0]CASE II: [/color]$p\neq 5$
Now let $F_0=0, F_1=1$ and $F_n$ be the $n^{th}$ Fibonacci number. It is easy to show that
$$x_{n+2}=F_{n+1}a+F_nb$$
[color=#00f]CLAIM 1.[/color] $F_p\equiv 1\pmod p$
Proof.
Using the general term formular for Fibonacci sequence,
\begin{align*}
F_p&=\frac{1}{\sqrt 5}\left(\left(\frac{1+\sqrt 5}{2}\right)^p-\left(\frac{1-\sqrt{5}}{2}\right)^p\right)\\
&=\frac{2}{2^p\sqrt 5}(\sum_{i=0}^{\frac{p-1}{2}}\sqrt 5\binom{p}{2i+1}5^i)\\
&\equiv\frac{1}{2^{p-1}}5^{\frac{p-1}{2}}\\
&\equiv5^{\frac{p-1}{2}} \pmod{p}
\end{align*}
The first equivalence follows from $p|\binom{p}{2i+1}$ for all $1\leq i\leq \frac{p-3}{2}$The last equvialence follow from $p\neq 2$ and Fermat little theorem. Now notice that $p|4k^2-5$, hence by Euler's criterion we have
$$5^{\frac{p-1}{2}}\equiv \left(\frac{5}{p}\right)=1\pmod p$$
This proves CLAIM 1. $\blacksquare$
[color=#00f]CLAIM 2.[/color] $F_{p-1}\equiv 0\pmod p$
Proof. The proof is similar to CLAIM 1. Again, using the general term formula for Fibonacci sequence,
\begin{align*}
F_{p-1}&=\frac{1}{\sqrt 5}\left(\left(\frac{1+\sqrt 5}{2}\right)^{p-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{p-1}\right)\\
&=\frac{1}{2^{p-2}}\sum_{i=1}^{\frac{p-3}{2}}\binom{p-1}{2i+1}5^i
\end{align*}
Notice that
$$\binom{p-1}{2i+1}=\frac{(p-1)!}{(2i+1)!(p-2i+1)!}\equiv \frac{(p-1)!}{(-1)^{2i+1}(p-1)!}\equiv -1\pmod p$$
Hence
$$F_{p-1}\equiv -\frac{1}{2^{p-2}}\sum_{i=1}^{\frac{p-3}{2}}5^i=-\frac{5^{\frac{p-1}{2}}-1}{4\cdot 2^{p-2}}\equiv 0\pmod p$$ $\blacksquare$
Now this implies that $x_{n+p-1}\equiv x_n \pmod p$ for all $n\in\mathbb N$. We now take $a=1$. It suffices to find $b$ which satisfies
$$F_{i+1}+F_ib\not\equiv 0\pmod p$$
for all $0\leq i\leq p-2$. Notice that there exists at most $1$ $b$ which satisfies
$$F_{i+1}+F_ib\equiv 0\pmod p$$
Since there are only $p-1$ incongruences, each of which eliminates at most $1$ chocie of $b$, together with the fact that there are $p$ residue classes modulo $p$, which gives $p$ choices for $b$, we conclude that we can always pick such $p$. This proves the lemma. $\blacksquare$
Now let $p_1,...,p_n$ be the prime factors of $m$. Then from the lemma for each $i$ we can pick $a_i$ and $b_i$ so that the sequence defined by $x_0=a_i$,  $x_1=b_i$ and
$$x_{n+2}=x_{n+1}+x_n$$
has all its terms relatively prime to $p_i$. Now by Chinese Remanider Theorem there exists integers $a$ and $b$ satisfying
$$a\equiv a_i\pmod{p_i}$$
$$b\equiv b_i\pmod{p_i}$$
Now by taking this $a$ and $b$ the proof is completed.

By CRT, it suffices to show the problem for $p\mid 4k^2-5$ where $p$ is a prime. Note that
\[x_n = F_{n-1}a + F_n b,\]
and since $(\tfrac{5}{p})=1$, we have that $F_{n+p-1}\equiv F_n\pmod{p}$ by Binet's thereom and Fermat's little theorem.

Let $t$ be the smallest positive integer such that $p\mid F_t$, so $t$ is the order of $\tfrac{1+\sqrt{5}}{1-\sqrt{5}}$ mod $p$, so $t\mid p-1$. Now, if $t\nmid n$, then $\tfrac{F_{n-1}}{F_n}$ can take at most $t-1$ values (since it repeats mod $t$), so it takes at most $p-2$ values, so there is some nonzero value mod $p$ it never achieves. Let this value be $c$.

Take $a=1$, $b\equiv -c\pmod{p}$. Then,
\[x_n = F_{n-1}a + F_n b\equiv F_{n-1}-F_nc\pmod{p}.\]
If $t\mid n$, then $F_n\equiv 0\pmod{p}$ and $F_{n-1}\not\equiv 0\pmod{p}$ (else $p\mid F_m$ for all $m$ which is not true), so $p\nmid x_n$. If $t\nmid n$, then $F_n\not\equiv 0\pmod{p}$, so this is never $0$ by the selection of $c$. Thus, these values of $a$ and $b$ yield all $x_n$ relatively prime to $p$, as desired.

Note that \[x_n = F_{n-1}a + F_n b,\]
[b][color=#f00]Claim:[/color][/b] There is a natural number $b$ such that $b^2=b+1 \pmod{m}$
[b][color=#f00]Proof:[/color][/b] We need $(2b-1)^2 \equiv 5 \pmod m$ i.e $5$ is a quadratic residue modulo $m$
Suppose that $m=\prod_{i=1}^{x} {p_i}^{e_i}$
By Hensel's Lemma,we can assure that each congruence has a solution and we can conclude by CRT.(here $f'(x)=8b-4$,and clearly we can choose such an $b$ s.t $ p \nmid 8b-4$ to get a solution.)
Now choose $a=1$ and clearly $x^n \equiv b^n \neq 0 \pmod m$ and $\gcd(x_n,m)=1$(by induction.

[hide=Solution]
By CRT, it suffices to show that for any prime $p$ dividing $m$, we can choose $a,b$ such that all the terms of $(x_n)$ are coprime with $p$.
If $5 \mid m$, choose $a \equiv 1\pmod{5} , b \equiv 3\pmod{5}$, one can check that this works.
Now let $p > 5$ be a prime dividing $m$. (Clearly $p$ cannot be $2$ or $3$.)
Since $5$ is a quadratic residue, we obtain $p \equiv 1,4\pmod{5}$.

Now define $F_0 = 0, F_1 = 1, F_2 = 1, F_{n+2} = F_{n+1} + F_n$ for $n \geq 1$. Then we have $x_n = F_{n-1}a + F_nb$ for all $n \geq 1$.
[color=#000]Lemma:[/color] For a prime $p \equiv 1,4\pmod{5}$, we have $F_p \equiv F_{p+1} \equiv 1 \pmod{p}$.
[hide=Proof:]
It is known that $F_{p}=\frac{\left(\frac{\sqrt{5}+1}{2}\right)^{p}-\left(\frac{1-\sqrt{5}}{2}\right)^{p}}{\sqrt{5}}$.
So, $F_{p} \equiv \frac{1}{2^p\sqrt{5}} \cdot ((1+\sqrt{5})^p - (1-\sqrt{5})^p) \equiv \frac{1}{2^{p-1}} \sum_{i=0}^{\frac{p-1}{2}} \binom{p}{2i+1}5^i \equiv 5^{\frac{p-1}{2}} \equiv 1 \pmod{p}$, since $5$ is a quadratic residue modulo $p$.
One can similarly show that $F_{p+1} \equiv 1 \pmod{p}$. So we are done. $\blacksquare$
[/hide]
From the lemma we have $F_{p-1} \equiv 0 \pmod{p} , F_p \equiv F_{p+1} \equiv 1 \pmod{p}$.
Now define $y_n = \frac{F_{n+1}}{F_n}$ for $F_n \not \equiv 0 \pmod{p}$. From the lemma it is clear that $y_n$ is periodic modulo $p$ and that period is at most $p-1$.
So, there exists $k \in \{1,2,\dots, p-1\}$ such that $k$ is not in the set $S=\{y_n \pmod{p} : 1 \leq n \leq p-1\}$.
Now choose $a,b$ such that $\frac{a}{b} \equiv -k \pmod{p}$.
Assume that there exists some $m$ such that $x_m \equiv 0 \pmod{p}$.
This means, $x_m \equiv F_{m-1}a + F_mb \equiv 0 \pmod{p} \implies -k \equiv \frac{a}{b} \equiv -\frac{F_m}{F_{m-1}} \pmod{p}$. But that contradicts the definition of $k$. So all the terms of the sequence $(x_n)$ are coprime to $p$, and we are done. $\square$

Is this right? This seems like a troll solution.

It's sufficient to construct a sequence mod all primes $p$ dividing $4k^2-5$, then combine them using CRT. If $p \mid 4k^2-5$, then $p$ is odd and $5$ is a quadratic residue $\!\!\!\mod p$. Let $a=1$, and choose $b$ such that $(2b-1)^2 \equiv 5 \pmod{p}$. It's easy to check that this condition forces $\frac{x_4}{x_3}=\frac{3b+2}{2b+1}=b=\frac{x_1}{x_0} \pmod{p}$ if $x_3$ is nonzero $\!\!\!\mod{p}$. Now, we check that $x_1,x_2,x_3 \not\equiv 0 \pmod{p}$:

If $x_1=b \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 1 \equiv 5 \pmod{p}$, so $p=2$, a contradiction.
If $x_2=b+1 \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 9 \equiv 5 \pmod{p}$, so $p=2$, a contradiction.
If $x_3=2b+1 \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 4 \equiv 5 \pmod{p}$, so there are no choices of $p$, a contradiction.

Since $\frac{x_3}{x_0} \equiv \frac{x_4}{x_1} \pmod{p}$, we know that $\frac{x_{n+3}}{x_n} \equiv \frac{x_3}{x_0} \pmod{p}$. Therefore, we can write $x_n=x_i \cdot \left(\frac{x_3}{x_0}\right)^{\left\lfloor \frac{n}{3} \right\rfloor}$, where $i \in \{0,1,2\}$. This is nonzero $\!\!\!\mod{p}$, so we are done.

Let $p$ be a prime and $p|m$. Note $p$ is odd.

Let $a\equiv 1\pmod{p}$ and $b\equiv k+2^{-1}\pmod{p}$. Assume towards a contradiction $b\equiv 0\pmod{p}$. Then, $2k\equiv -1\pmod{p}$. Then, $5\equiv 4k^2\equiv 1\pmod{p}$, so $p|4$, so $p=2$, a contradiction.

We have $$b^2\equiv k^2+k+2^{-2}\equiv 2^{-2}(4k^2+4k+1)\equiv 2^{-2}(5+4k+1)\equiv 2^{-1}(2k+3)\equiv k+2^{-1}+1\equiv b+1\pmod{p}.$$

We claim $x_n\equiv b^n\pmod{p}$. We will prove it with induction with $n=0, 1$ as the base case. We have
$$x_n\equiv x_{n-1}+x_{n-2}\equiv b^{n-1}+b^{n-2}\equiv b^{n-2}(b+1)\equiv b^n\pmod{p}.$$
So, since $b\not \equiv 0\pmod{p}$, none of the $x_n$ are equivalent to $0\pmod{p}$. Thus the whole sequence is relatively prime for $p$.

So, for each prime $p|m$, there exists a residue $b_p$ such that when $x_0\equiv 1\pmod{p}$ and $x_1\equiv b_p\pmod{p}$, $x_i$ is never a multiple of $p$. By the Chinese Remainder Theorem, there exists $B$ such that $B\equiv b_p\pmod{p}$ for all $p|m$. When $a=1$ and $b=B$, the sequence will always be relatively prime to $m$.

Ok so the motivation of the above solution comes from the fact that for each prime $p$ dividing $m$ we can write $\sqrt{5} \equiv 2k$ and the recursion can be characterized as usual in the form
$$x_n \equiv x\left(\frac{1+2k}{2}\right)^n+y\left(\frac{1-2k}{2}\right)^n$$
and it clearly necessary that all of these should be nonzero mod $p$ (in fact by CRT it is also sufficient, so we've reduced the problem to this). The best way to deal with this is to do the stupidest thing possible and set $y=0$ which yields (after adding $4k^2-5$) the above solution.

Note that $5$ is always a quadratic residue of any $p\mid m$. Let $(2k-1)^2\equiv 5\pmod p$ then $p\mid k^2-k-1$. Then pick $a=1$, $b\equiv k\pmod p$, and we see that $x_n\equiv k^n\pmod p$ and by CRT we're done.

We show that for each prime divisor $p$ of $m$, we can find some $a_p$ and $b_p$ so that $\left(x_n\right)$ has all its terms nonzero modulo $p$. Let $p$ be a prime divisor of $m$.

If $5$ is a prime divisor of $m$ and $p=5$, we take $a_5=1$ and $b_5=3$. Then $\left(x_n\right)$ repeats $1, 3, 4, 2, 1, 3,\dots$ modulo $5$, with all terms being nonzero.

Now we consider $p\neq 5$. Then $k\not\equiv 0\pmod{p}$, since that would otherwise imply $p\mid 5$. We also note that $p$ is odd, since $2$ does not divide $4k^2-5$. Now let $\alpha\equiv \frac{1+2k}{2}$ [hide=motivation] $\left(x_n\right)$ is a linear recurrence with characteristic polynomial $x^2-x-1$; the roots of this polynomial are $\frac{1\pm\sqrt{5}}{2}$, and $\sqrt{5}\equiv \pm 2k$[/hide] We have
\[\alpha^2\equiv \frac{1+4k+4k^2}{4}\equiv \frac{6+4k}{4}\equiv \frac{3+2k}{2}\equiv \alpha + 1\]
This also implies $\alpha\not\equiv 0$. Now, let $a_p\equiv 1$ and $b_p\equiv \alpha$. We claim by induction that $x_n\equiv \alpha^n$. We have the base cases $n=0$ and $n=1$ already. Then for the induction step, we have for $n\geq 2$ that
\[x_n\equiv x_{n-1}+x_{n-2}\equiv \alpha^{n-1}+\alpha^{n-2}\equiv \alpha^{n-2}(\alpha+1)\equiv \alpha^{n-2}\alpha^2\equiv \alpha^n\]
which completes the induction. Then, since $\alpha\not\equiv 0$, all terms of $\left(x_n\right)$ are nonzero modulo $p$.

Using the Chinese Remainder Theorem, we can guarantee the existence of $a$ and $b$ such that $a\equiv a_p\pmod{p}$ and $b\equiv b_p\pmod{p}$ for all prime divisors $p$ of $m$. This guarantees that $\left(x_n\right)$ will have all its terms not divisible by any prime divisor of $m$, and thus relatively prime to $m$.

An alternative continuation after the
$x_n = x(\frac{1+k}{2})^n + y(\frac{1-k}{2})^n$ construction:

There are at most $p-1$ possible unique modulo $p$ pairs for $((\frac{1+k}{2})^n,(\frac{1-k}{2})^n)$
Each pair eliminates at most $p$ different potential pairs $(x,y)$
We are done as there must be at least one pair leftover


function
algebra
polynomial
superior algebra
superior algebra theorems

How does the polynomial in $ p$ given by $ (p^5\minus{}u)^2\equal{}u'^2s^2$ have coefficients that are all symmetric functions here? $ u$ and $ u'$ and $ s$ are functions. If you swap these functions with each other you get a different polynomial!

actually: symmetric polynomials,when you swtich the coefficients with each other the polynomial is unchanged. $ x_1 x_2 \plus{} x_3 x_2$ is symmetric for example

I know what a symmetric function is. You simply did not specify what the variables are for which it is supposed to be symmetric.

E.g. $ x^2 \plus{} y \plus{} z$ is symmetric in respect to $ y,z$ but not in respect to $ x,y$.

Aha, every coefficient of $ p$ should be symmetric. I don't think that we should worry about those things on the rt. side of the eq. In this case it seems that the polynomial is not symmetric w.r.t. those coefficients from looking. But it ought to be.

[quote="Los"]Aha, every coefficient of $ p$ should be symmetric.[/quote]

The expansion in $ p$ is given by

$ p^{10} \minus{} 10u p^5 \plus{} (u^2 \minus{} u'^2 s^2)$

and those coefficients aren't symmetric in any pair of variables, so I'm not sure what you're talking about.  It sounds to me from your past several posts that you're working from a textbook of some sort; care to post the relevant exercise?

It is from Abel's paper on the solvability of polyomials of general form of deg. 4 or less.

Perhaps something else is meant by "symmetric function."  If $ u, u', s$ are really functions (presumably of some other variables; perhaps the roots of a quintic?), Abel might mean that $ u, u', s$ are symmetric functions of those variables.


Has anyone read [b]Romance of the Three kindoms [/b]before??? ... You should, another great neurology book to check out is [b]Phantoms in the Brain[/b]

no on cares???

*bump*

more ppl need to read...lol

I haven't read either of those... but have you read the [b]His Dark Materials[/b] trilogy? Other stuff I like is books on which movies are based, i.e. Jurassic Park. Of course, I like Tolkein, C.S. Lewis, and most of the other really, [b]really[/b] famous authors. Like probably everybody else here, I read really fast. My school librarian makes cracks about how I'll check out a book for an afternoon. Funny thing is, she's correct... I really do read that fast...

-Egotistically, Isaac

His dark materials....GREAT SERIES...FOR  EVRY1...READ IT NOW!!!!

allright sry for caps...if u need a challenge (or just a long book) try the Robert Jordan books...

I've heard that it's pretty good. Another great series- Foundation. I also liked the Hitchhiker's Guide to the Galaxy "trilogy".

the hitchhiker's guide series rocked..........................well i read more mathbooks than novels..........but that most likely applies to everyone here.............................but fantasy rocks..........i read a bit of foundation before stopping for some reason that i dont remember.............or in other words G4 classified..................i dunno what that means.........but that came from Rush Hour...........good movie...............watch it in your spare time or somat...................................it goes without saying though that the lotr rules all...........except me........looooooooool.........woooooooooooooot.............of course though................the art of problem solving books are actually the best books..........hehe........... :D

Mods, could you please delete all the "wooooooooooooooooooot!"'s and the "..........................................."'s from Syntax's posts? They're really hard to read... And Syntax, could you stop that? Please?

That said, I have no clue how I could have forgotten to mention the Orson Scott Card books, neh? As in the Ender and Bean series? Who else thinks that those rank among the top 5 science fiction books / series?

-Isaac

Yah, Syntax is like that crazy now, although he wasn't this hyper before school ended... At least I didn't think he was.

Anyways, I like Lord of the Rings, and Dune. And then there's mom that makes me read all sorts of books that I have no interests in whatsoever. Has anyone read Dune or its prequels or sequels?

fine i'll stop. but i just dont like thinking when to end a sentence. yes, im that lazy. i even thought about not putting spaces in between words. but that would be a bit too harsh. fine i aquisce only because well, i dont quite know but fine.

Thank you Sytnax...
I am currently reading Foundation and Earth. I've paused in my reading of the Dune sextet. I've read the first three, but Heretics is too boring at the beginning... And of course I like Lord of the Rings. That's definitely in the top 5 fantasy...

-Isaac

I agree that Orson Scott Card's Ender and Bean books are some of the greatest books.  His Pastwatch: the redemption of Christopher Columbus is pretty good too.
Some other books that I like are the Left Behind books.

Romance of the Three Kingdoms is great...any Chinese swordfighting novel fans out here? None? Darn. How about Austen? I love LoTR. :D

i find it fun and entertaining to read some berenstain (mind the spelling) bears books for the fun of it at night at 4 a.m. wondering what else to do to pass time until sunrise

And just [i]how[/i] old are you, Syntax?

-Isaac

You like those [i]bears?[/i] The Arthur books are [b]so[/b] much better. (Especially the one where he switches his nose. It's so exciting! And there's such a dramatic climax!)[/i]

mine was a joke, but i think you need some psychological help. nuff said

Yah, I found Heretics, Children of Dune, God Emperor all boring, but then the 5th and 6th are exciting (to me that is).

Sum1 read Romance of the Three Kingdoms??? IT ROX0RED...(pardon my caps)....any one else read it?

I haven't read the Romance of the Three Kingdoms.  Does it have another name? (I can't seem to find it at the library).

anybody read David Eddings?  the Belgariad and Mallorean series are really awesome
 i haven't read any Card but everybody tells me that i should read the Enders

how about Redwall? I've read all of those...they're pretty good

then theres [i]I, Jedi[/i] which is pretty much the greatest book of all time
-it should probably be required reading in school
its by Michael A. Stackpole

i admit that i am a complete failure as a person because i have not read lotr  :oops:   - i have read the hobbit though
i have also read Dune but haven't got around to any of the sequels yet

when i have like ten years of free time sometime i might try the Wheel of Time

im surprised no one including me has said this yet. the wheel of time series. also umm. whats it called. the dragonlance chronicles thingie. the first book is dragons of autumn twilight. ajods

LOTR LOTR LOTR LOTR... :D
Anyone here read Silmarillion?

It's interesting as to how homogeneous we are here- I'm no different, and enjoyed the foundation series as well as the hitchiker's guide. Some other books worth reading are The Fountainhead by Ayn Rand (my namesake) and of course the earlier Tom Clancy books- Hunt for Red October and such.

The Fountainhead was the best of Rand's stories...Atlas Shrugged I found a little tedious...

A Canticle For Leibowitz, 1984, Brave New World, anything like that...

1984 is the best book in the universe. orwell is the god. please read signature if you do not understand me.

ayn rand... dont read anthem. it gets really confusing whether or not they are talking about we (as in me and another person) or we (as in I CANT TALK ABOUT MYSELF IN THE FIRST PERSON!). grness.

right now im reading when the wind blows, by james patterson... seems pretty good. girl with wings+veterinarian=what more could a person want?

redwall is the best. martin the warrior is the coolest of redwall (of course, someone stole my copy of that one, grness!)

the only thing i have to say about the hitchhiker's series: "if i ever meet myself, i'm going to hit myself so hard i wont know what hit me!"

oh and if ANYONE tells you that flatland is a good book, tell them to get a life and get out of the 19th century. yeah, its math oriented, woot woot, but come on. there are better math-fiction books out there! that arent sexist!

[quote="bleumoose"]oh and if ANYONE tells you that flatland is a good book, tell them to get a life and get out of the 19th century. yeah, its math oriented, woot woot, but come on. there are better math-fiction books out there! that arent sexist![/quote]

Well, you have to grant Abbott this: he didn't know any better.  That was a rather, uh, common belief back then.

But yeah, it is - to say the least - old, musty reading.  I prefer exciting books, and not necessarily science fiction, ether.  Hitchhiker's was okay... very random (like me) but repeating the same thing over (not like me)(not like me)(not like me).  I mean, if you wanna make your books so silly that it'll drive the readers stupid, then I have no problem with that.  I just don't like books that drive your readers stupid by repeating the [i]same[/i] silly things, like how mice are the smartest things on earth, and how Earth is a big computer.

Foundation was much better.  Had some nice plot twists, stuff you wouldn't expect, and I was impressed at how Asimov thought a lot about the politics of Foundation with respect to the universe, and how everything seemed to rest on the point of a sword - tip it one way just a little and the whole thing comes crashing down.

And Ender's Series was good.  Xenocide was very deep, but good in that way.  The three [i]Shadow[/i] Ender books were good, but I've heard a lot of people say they were pointless.  Ah, well.  So they threw Ender out of the picture and picked up his "Shadow" (Bean - who's now twice his height).  So live with it!  You don't have to depend on a classic character in order to have an interesting read.

Gotta get back to reading.  Heh, heh.

-mathfanatic

The Foundation and Ender series are great, but I especially likethe murder mysteries written by Tony Hillerman.

Also, I have read the all five books in the Lord of the Ring series.

The Hobbit and Sil don't count as part of the series. But they are sstill awesome  :D

-Isaac

[color=white]I agree, Issac.

Just responding to bookworm...somebody in my math class last year took down a bunch of funny things said during class (it was [i]very[/i] lighthearted) and one of them was as follows:[/color]
[quote="N.K"]War is not the answer...it is the solution.[/quote]
[color=white]You most likely would not agree.  :lol: [/color]

I've always thought that it went:

War is not the answer, it is the question, and the answer is yes.

Not that I agree.

Not war...
Violence is not the answer! Violence is the question, and the answer is yes.

-Isaac, NOT Issac

I just finished a really great book called [u]Mapping Human History[/u] by Steve Olson. It's about genetic anthropology - using genetics to tell us things about human history, race, migration, and even language. It has a lot of surprising claims, like why Confucious, Plato, and even Ghengis Khan are ancestors of every person alive today; how many people there have likely ever been on Earth; and the evidence for a "Mitochondrial Eve."

It has a definite presupposition of believe in Evolution, so if you're not in that camp it may not do much for you (then again, it might still be interesting). But it's one of the most fascinatings books that I've read in a while.

I read Mapping Human History a few months ago and also found it fascinating.  It was a finalist for the National Book Award.  I met Steve at National MATHCOUNTS in 2002 - his next book, which will be released next spring, will be about a recent IMO team.  If there's interest among the people on this board, Steve may be available for a 'Math Jam' to talk about his upcoming book & perhaps talk a bit about nonfiction writing in science & math.  Given his success and quality of writing, those of you interested in writing may find it very interesting.  Let me know if you are interested & I'll ask Steve.

Awesome.  I would love to attend a Math Jam by him.

I'll see what I can do.  It likely wouldn't be until late spring, after the book is published.

I loved the Ender's Game series by Orson Scott Card.. anyone else read that?
and then, as for current bestsellers, of course.. there's Dan Brown's The Da Vinci Code (really GOOD!) and The Five People You Meet in Heaven by Mitch Albom (good book too=D) and I'm waiting to read Pompeii.. but of course there's like a five mile long waiting list at the library.. =(

[quote="MathFiend"]If you want to read something that makes you have to think a lot...try "The Universe in a Nutshell" bu Stephen Hawking...I recently started it and thoroughly confused and am only on page 33...so, it makes you think a lot if anyone really cares...[/quote]

Have you read his book [u]A Brief History of Time[/u]? It explains some of the general things about cosmology in a really nice way. It doesn't contain much about string theory and M-Theory though, because it was written before those fields really took off.

Another good laymen's book on cosmology is one called [u]Frozen Star[/u] by George Greenstein. I read it when I was young, so it's kind of old now and some of the information in it is a bit outdated. But it's explainations of things like neutron stars, pulsars, and black holes are excellent!

If you're interested in more of this kind of stuff you should start a thread in the Round Table about Cosmology. It could make an interesting discussion.

no, I haven't read A Brief History of Time yet...it was checked out of the library when I was there, so I got The Universe in a Nutshell...

"Island of the Aunts" is good, like all books by Eva Ibbotson.  She's a really good author.  "Phantom Tollbooth" is funny and has some math puns (yay!) but that's not the entire point of the book, which is cool.  Artimis Fowl, Dealing with Dragons, and Series of Unfortunate Events: Series.  "The Illustrated Man" is a really weird book, science fiction, by Ray Bradbury, kinda cool though.  The Left Behind series is Christian series about the Rapture, it's really cool, but NEVER try the "teenager" ones.  I swear, if those are teenage books, teenagers must be REALLY stupid.  The adult ones are better.  One last series, Wayside School.  Okay, it's for younger kids, but so what?  I  like them anyway!

[quote="lunmu"]I loved the Ender's Game series by Orson Scott Card.. anyone else read that?[/quote]

I'm willing to bet that a lot of people here have read Orson Scott Card.  If I recall correctly, Ender's Game was really popular at Mathcounts nationals, but that's somewhat surprising, because I would have expected everyone to already have read that book...

Have you read the other Ender books?  Or, say, the books about Bean?  Ender's Shadow was great.  Shadow of the Hegemon was okay.  Shadow Puppets was decent.  I wonder what the next one will be like...

Oh, and there are a ton of other books by Orson Scott Card.  I read the Homecoming books (about how the earth is almost destroyed in a nuclear war, so the survivors are taken to a distant planet and watched over by this supercomputer, but the supercomputer is dying...).  Actually, I only read 4 and a half of them; the fifth one was sooo boring that I stopped halfway.  Also, the Speaker for the Dead/Xenocide/Children of the Mind books are good.

Hm.  I think I'll make a poll about Orson Scott Card.


abstract algebra
group theory
superior algebra
superior algebra unsolved

I would like to find all short exact sequences of abelian groups $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ where $M$ has at most 9 elements.

I have managed to write down 12 of them but I'm not sure whether they are correct and whether there are more.

OK, repost after three Edits:

The sequence is uniquely defined by $M$, $M'$ and the injection since $M''$ will be isomorphic to the quotient of $M$ and $\mathrm{Im}(M')$ (with the canonical mapping, we get others by adding an automorphism of $M''$).
So we just have to choose $M$ and then a subgroup $M'$, then a automorphism of the quotient.

For $M$, there are the following possibilities:
- $C_{m}: = \mathbb{Z}/m\mathbb{Z}$, for $m=1,2,3,...,9$.
- $C_{2}^{2}, C_{2}^{3}, C_{2}\times C_{4}, C_{3}^{2}$.

This are already more than $12$ groups, namely $13$ of them.

$C_{m}$ has one subgroup for each divisor of $m$, $C_{2}^{2}$ has $5$ subgroups, $C_{2}^{3}$ has $16$, $C_{2}\times C_{4}$ has $8$, $C_{3}^{2}$ has $6$.
That gives in total: $58$ subgroups and at least that many sequences...

So the automorphisms of the quotients are to be counted also, but thats a rather lengthy task...

There are $\sum_{n=1}^{9}\sum_{d|n}\varphi(d) = \sum_{n=1}^{9}n = 45$ sequences with $M$ cyclic.

There are $10$ with $M=C_{2}^{2}$, $218$ with $M=C_{2}^{3}$ and $51$ for $M=C_{3}^{2}$.

Remaining case is $C_{2}\times C_{4}$, which gives $22$ (the latter done, not like the ones before, by stupid counting).

Total number of sequences: $346$, about $29$ times what you said  :lol:


[quote="Khanh_Vo"][quote="April"]Problem 1. http://www.mathlinks.ro/viewtopic.php?t=217565
Problem 2. http://www.mathlinks.ro/viewtopic.php?t=217569
Problem 3. http://www.mathlinks.ro/viewtopic.php?t=217573
Problem 4. http://www.mathlinks.ro/viewtopic.php?t=217592
Problem 5. http://www.mathlinks.ro/viewtopic.php?t=217595
Problem 6. http://www.mathlinks.ro/viewtopic.php?t=217598[/quote]

Do you have Vietnam TST 1989? I really need it.[/quote]
search it by http://www.google.com

I think your advice is useless. [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=5820]N.T.TUAN[/url] promised to post some missing [url=http://www.mathlinks.ro/resources.php?c=186&cid=41]Vietnam TSTs[/url] but so far he is very lazy...


I would write out the entire solution, but it would get messy since I don't know the codes, however, using the fact that to get the reciprocal of the fraction, you switch the bases of the logs, thus the sequence is all with base of $100!$, thus you can multiply everything together so you will realize that it would be $100(99)(98)....(1)$ which is the same as $100!$ so it would be $1$.

[quote="xxreddevilzxx"]this was too easy to be in Intermediate. At least i think so.[/quote]
Oh well, it's always good to brush up on these things...the more you learn, the more you forget ;)


Let $ m_{a}, m_{b}$ and $ m_{c}$ the medians of the triangle $ ABC$.
Show that $ (m_{a} \plus{} m_{b} \plus{} m_{c})^{2}\leq\frac {(a \plus{} b \plus{} c)^{2} \plus{} 6 (a^{2} \plus{} b^{2} \plus{} c^{2})}{4}$

Consider medians $ m_{a}$ and $ m_{b}$ in the triangle $ ABC$. By Ptolemy's inequality in quadrilateral $ ABA'B'$, we have:

$ AA' * BB'\leq\ AB*A'B' \plus{} A'B*AB'$

Which is equivalent to:

$ m_{a}m_{b}\leq\frac {c^2}{2} \plus{} \frac {ab}{4}$

In a similar way:

$ m_{b}m_{c}\leq\frac {a^2}{2} \plus{} \frac {bc}{4}$

$ m_{c}m_{a}\leq\frac {b^2}{2} \plus{} \frac {ca}{4}$

Adding up these inequalities we have:
$ m_{a}m_{b} \plus{} m_{b}m_{c} \plus{} m_{c}m_{a}\leq\frac {a^2}{2} \plus{} \frac {b^2}{2} \plus{} \frac {c^2}{2} \plus{} \frac {ab}{4} \plus{} \frac {bc}{4} \plus{} \frac {ca}{4}$

Thus:
$ 2(m_{a}m_{b} \plus{} m_{b}m_{c} \plus{} m_{c}m_{a})\leq(a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {1}{2}(ab \plus{} bc \plus{} ac)$ (*)

But we also have:

$ m_{a}^2 \equal{} \frac {1}{4}(2b^2 \plus{} 2c^2 \minus{} a^2)$

$ m_{b}^2 \equal{} \frac {1}{4}(2c^2 \plus{} 2a^2 \minus{} b^2)$

$ m_{c}^2 \equal{} \frac {1}{4}(2a^2 \plus{} 2b^2 \minus{} c^2)$

Thus:
$ m_{a}^2 \plus{} m_{b}^2 \plus{} m_{c}^2 \equal{} \frac {3}{4}(a^2 \plus{} b^2 \plus{} c^2)$

Therefore, using (*):
$ (m_{a} \plus{} m_{b} \plus{} m_{c})^2 \equal{} (m_{a}^2 \plus{} m_{b}^2 \plus{} m_{c}^2) \plus{} 2(m_{a}m_{b} \plus{} m_{b}m_{c} \plus{} m_{c}m_{a})\leq$

$ \frac {3}{4}(a^2 \plus{} b^2 \plus{} c^2) \plus{} (a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {1}{2}(ab \plus{} bc \plus{} ac) \equal{} \frac {(a \plus{} b \plus{} c)^{2} \plus{} 6 (a^{2} \plus{} b^{2} \plus{} c^{2})}{4}$


geometry
geometric transformation
rotation
calculus
integration
LaTeX

Here's a neat problem I had to solve for my mechanics class 2 years ago:

Nine square holes have been cut in a flat square plat  The plate has edge length $ L$ and the holes have edge length $ a$.  The holes are located at the centers of the small squares formed by dividing each side of the square into three equal sections.  Find the rotational inertia for rotations about an axis perpendicular to the plane passing through its center.

[hide="Hint"]Parallel-axis theorem[/hide]

Here's my attempt.

moment of total= moment of square- moment of small square holes

The closest square holes (4 total) will have moments of:
M (a/L)^2 * [(L^2)/9 + (a^2)/6]= call this I1

The farthest holes (4 total) will ahve moments of  (I2):
M (a/L)^2 * [2(L^2)/9+ (a^2)/6]= I2

The center holes has a moment of I3= M (a/L)^2 * ((a^2)/6)

So the moment of inertia = M(L^2)/6- I3- 4I2-4I1

Is that the right way?  I basically looked up the moment of inertia, but is there a way to integrate and show that the moment of inertia of a square is M L^2/6 from I= Integral r squared *dm?

Thanks. :)

[quote="integration"]Are they square holes?
[/quote]
Yes they are.  I can't fully grasp your solution because it's not in $ \text{\TeX}$ so it's hard to read.  To turn it into $ \text{\TeX}$, just add dollar signs around your math and for fractions just use[code]\frac{numerator}{denominator}[/code]
And you don't need a * for multiplication.  So your first line would be
$ M\left(\frac{a}{L}\right)^{2}\left[\frac{L^2}{9}+\frac{a^2}{6}\right]=I_{1}$
You can click on it to see the syntax.  Once you convert the rest of your solution to $ \text{\TeX}$, I'll be able to tell you if it's correct.

M\left(\frac{a}{L}\right)^{2}\left[{2}\frac{L^2}{9}+\frac{a^2}{6}\right]=I_{2}

M\left(\frac{a}{L}\right)^{2}\left[\frac{a^2}{6}\]=I_{3}

Just to test...  I still don't get how you guys use latex... :oops:

I'll try to post later when I can use it properly.

You've almost got the $ \text{\LaTeX}$ down.  Just remember to surround the math with $ \$$.  Here's what you've got:
$ M\left(\frac{a}{L}\right)^{2}\left[{2}\frac{L^2}{9}+\frac{a^2}{6}\right]=I_{2}$

$ M\left(\frac{a}{L}\right)^{2}\left[\frac{a^2}{6}\right]=I_{3}$,
which is pretty good.  I'll look over the actual solution once it's complete.


$ (a)$Find infinetily many pairs of integers a and b with 1<a<b,$ 1<a<b$so that$ ab$exactly divides$ a^{2}\plus{}b^{2}\minus{}1.$
$ (b)$With a and b as above, what are the possible values of$ \frac{a^{2}\plus{}b^{2}\minus{}1}{ab}$

[quote="yildobekir"]$ (a)$Find infinetily many pairs of integers a and b with 1<a<b,$ 1 < a < b$so that$ ab$exactly divides$ a^{2} \plus{} b^{2} \minus{} 1.$
$ (b)$With a and b as above, what are the possible values of$ \frac {a^{2} \plus{} b^{2} \minus{} 1}{ab}$[/quote]
[hide]
Let $ b \equal{} a \plus{} r$, with r an integer, because $ b$ is an integer greater than $ a$.

Then the fraction becomes:

$ \frac {(a^2) \plus{} (a^2 \plus{} 2ar \plus{} r^2) \minus{} 1}{a(a \plus{} r)} \equal{}$

$ \frac {2a^2 \plus{} 2ar \plus{} r^2 \minus{} 1}{a(a \plus{} r)}$.

$ (a \plus{} r)$ must divide the numerator.

After dividing $ (2a^2 \plus{} 2ar \plus{} r^2 \minus{} 1)$ by $ (a \plus{} r)$, the quotient is
$ 2a \plus{} 2(r \minus{} 1)$, with a remainder of $ (r^2 \minus{} 2r \plus{} 1)$.

The remainder is equal to $ (r \minus{} 1)^2$.

For $ (a \plus{} r)$ to divide the dividend, the remainder must equal zero.

So, $ (r \minus{} 1)^2 \equal{} 0$.

Then $ r \equal{} 1$.

Substituting $ r \equal{} 1$ into the fraction, it becomes:

$ \frac {2a^2 \plus{} 2a}{a(a \plus{} 1)}$.

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

Answer to a)

Let $ t \ge 2$, and $ t$ is an integer.

Let $ a \equal{} t$.
Then $ b \equal{} t \plus{} 1$.

Then $ (t, t \plus{} 1)$ represents an infinite number of solutions
for $ (a, b)$.

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

Answer to b)

If the numerator of the fraction is factored, and the fraction simplified,
it becomes:

$ \frac {2a(a \plus{} 1)}{a(a \plus{} 1)} \equal{} 2$.

Then the possible values of the given fraction consist of only $ 2$.
[/hide]


geometry
trigonometry
rectangle
perimeter
cyclic quadrilateral
number theory unsolved
number theory

[quote="Moonmathpi496"]Find all cyclic quadrilaterals having sides lengths odd integer and the area a prime number.

[hide="about the problem"][p.s. it is both geometry and number theory.but the solution is much more number theoretic][/hide][/quote]

Let $ ABCD$ the quadrilateral (in this order), with $ AB \equal{} a$, $ BC \equal{} b$, $ CD \equal{} c$, $ DA \equal{} d$ and $ \alpha \equal{}$internal angle$ (BA,BC)$=$ \pi\minus{}$internal angle$ (DC,DA)$ ($ A,B,C,D$ cocyclic).

We have $ AC^{2}\equal{} a^{2}\plus{}b^{2}\minus{}2ab\cos(\alpha)$ and $ AC^{2}\equal{} c^{2}\plus{}d^{2}\plus{}2cd\cos(\alpha)$ and so $ \cos(\alpha) \equal{}\frac{a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}}{ab\plus{}cd}$
We also have Area_of_ABC$ \equal{}\frac{ab\sin(\alpha)}{2}$ and Area_of_CDA$ \equal{}\frac{cd\sin(\pi\minus{}\alpha)}{2}$ $ \equal{}\frac{cd\sin(\alpha)}{2}$ and so the area $ S$ of the quadrilateral is $ S \equal{}\frac{ab\plus{}cd}{2}\sin(\alpha)$

So we have $ \cos(\alpha) \equal{}\frac{a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}}{ab\plus{}cd}$ and $ \sin(\alpha) \equal{}\frac{2S}{ab\plus{}cd}$ and so $ (\frac{a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}}{2})^{2}\plus{}S^{2}\equal{} (\frac{ab\plus{}cd}{2})^{2}$

Hence, for symetry reason, we have :
E1: $ (\frac{a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}}{2})^{2}\plus{}S^{2}\equal{} (\frac{ab\plus{}cd}{2})^{2}$
E2: $ (\frac{b^{2}\plus{}c^{2}\minus{}d^{2}\minus{}a^{2}}{2})^{2}\plus{}S^{2}\equal{} (\frac{bc\plus{}da}{2})^{2}$

Then :
1) Either one of the two first elements, WLOG say $ a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}$ is zero :
$ a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}\equal{} 0$ and $ ab\plus{}cd \equal{} 2S$
So we have $ 4S \equal{} (c\plus{}d)^{2}\minus{}(a\minus{}b)^{2}$ and, since $ a,b,c,d$ are odd and $ S$ is prime, $ c\plus{}d\plus{}|a\minus{}b| \equal{} 2S$ and $ c\plus{}d\minus{}|a\minus{}b| \equal{} 2$
But also $ 4S \equal{} (a\plus{}b)^{2}\minus{}(c\minus{}d)^{2}$ and $ a\plus{}b\plus{}|c\minus{}d| \equal{} 2S$ and $ a\plus{}b\minus{}|c\minus{}d| \equal{} 2$
And we have $ a\plus{}b \equal{} c\plus{}d \equal{} S\plus{}1$ and $ |a\minus{}b| \equal{} |c\minus{}d| \equal{} S\minus{}1$, so $ \max(a,b) \equal{}\max(c,d) \equal{} S$ and $ \min(a,b) \equal{}\min(c,d) \equal{} 1$ and $ \cos(\alpha) \equal{} 0$
This gives solutions like $ (S,1,1,S)$ or $ S,1,1,S)$:
Quadrilateral with successive sides $ p,1,p,1$ ($ p$ odd prime) with angle between 1-sides and p-sides equal to $ \frac{\pi}{2}$ : rectangles
Quadrilateral with successive sides $ p,1,1,p$ ($ p$ odd prime) with angle between 1-sides and p-sides equal to $ \frac{\pi}{2}$

2) either none of $ a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}$ or $ b^{2}\plus{}c^{2}\minus{}d^{2}\minus{}a^{2}$ is zero; Then we have two pythagorean triples.
$ S\neq 2$ since equation $ x^{2}\plus{}4 \equal{} y^{2}$ have just one solution $ x \equal{} 0$ and $ y \equal{} 2$ and we know $ x\neq 0$
So $ S$ is odd and it exists $ u,v,m,n$ such that :

$ a^{2}\plus{}b^{2}\minus{}c^{2}\minus{}d^{2}\equal{} 4uv$
$ S \equal{} u^{2}\minus{}v^{2}\equal{} (|u|\minus{}|v|) (|u|\plus{}|v|$
$ ab\plus{}cd \equal{} 2(u^{2}\plus{}v^{2})$

$ b^{2}\plus{}c^{2}\minus{}d^{2}\minus{}a^{2}\equal{} 4mn$
$ S \equal{} m^{2}\minus{}n^{2}\equal{} (|m|\minus{}|n|) (|m|\plus{}|n|$
$ bc\plus{}ad \equal{} 2(m^{2}\plus{}n^{2})$

But, since $ S$ is prime, we have $ |u| \equal{} |v|\plus{}1 \equal{} |m| \equal{} |n|\plus{}1$
So $ ab\plus{}cd \equal{} bc\plus{}ad$ and $ (a\minus{}c)(b\minus{}d) \equal{} 0$ and so WLOG $ a \equal{} c$

And this leads to $ a \equal{} b \equal{} c \equal{} d \equal{} 1$

==============================
So the only solutions are :
Quadrilateral with successive sides $ p,1,p,1$ ($ p$ odd prime) with angle between 1-sides and p-sides equal to $ \frac{\pi}{2}$ : rectangles
Quadrilateral with successive sides $ p,1,1,p$ ($ p$ odd prime) with angle between 1-sides and p-sides equal to $ \frac{\pi}{2}$

Area: $ p\equal{}\sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)}$, where $ s\equal{}\frac{1}{2}(a\plus{}b\plus{}c\plus{}d)$.
Now $ p^{2}\equal{}(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)$ and we have two cases.
1) $ s\minus{}a\equal{}s\minus{}b\equal{}s\minus{}c\equal{}1$, $ s\minus{}d\equal{}p^{2}$. Now $ a\equal{}b\equal{}c\equal{}s\minus{}1$, so $ a\plus{}b\plus{}c\equal{}3s\minus{}3$. But $ a\plus{}b\plus{}c\equal{}2s\minus{}d < 2s$, so $ 3s\minus{}3 < 2s$ implies  $ s<3$. But if $ s\equal{}2$ then $ s\minus{}d$ cannot be a square of prime.
2) $ s\minus{}a\equal{}s\minus{}b\equal{}1$, $ s\minus{}c\equal{}s\minus{}d\equal{}p$. Now $ a\equal{}b\equal{}s\minus{}1$, so $ a\plus{}b\equal{}2s\minus{}2$. But $ a\plus{}b\equal{}2s\minus{}c\minus{}d\le 2s\minus{}2$, so $ 2s\minus{}2\le 2s\minus{}2$ implies that $ c\equal{}d\equal{}1$. Now $ a\equal{}b\equal{}p$. The sides of quadrilateral are $ 1, 1, p, p$ (rectangle or deltoid).

[hide="Hint"]Use Brahmagupta's formula: If a cyclic quadrilateral with sides $ a,b,c,d$ has area $ p$ and semi-perimeter $ s$, then $ p^{2}\equal{} (s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)$.[/hide]

[hide="Solution"]If $ p$ is prime in the above relation, then $ p^{2}$ has only three distinct divisors namely $ 1,p$ and $ p^{2}$. But in our relation there are four factors, which implies two of them must be equal. WLOG suppose that $ s\minus{}a \equal{} s\minus{}b$. Also note that $ s\minus{}a \equal{} s\minus{}b \equal{} 1$ or $ p$. Let $ a \equal{} 2A\minus{}1,b \equal{} 2B\minus{}1,c \equal{} 2C\minus{}1,d \equal{} 2D\minus{}1$. Then we easily get $ s\minus{}a \equal{} B\plus{}C\plus{}D\minus{}A\minus{}1,s\minus{}b \equal{} C\plus{}D\plus{}A\minus{}B\minus{}1,s\minus{}c \equal{} D\plus{}A\plus{}B\minus{}C\minus{}1,$ and $ s\minus{}d \equal{} A\plus{}B\plus{}C\minus{}D\minus{}1$.

If $ s\minus{}a \equal{} s\minus{}b \equal{} 1$, then $ (s\minus{}c,s\minus{}d) \equal{} (1,p^{2}),(p,p)$. First let $ (s\minus{}c,s\minus{}d) \equal{} (1,p^{2})$. From this we get the system
$ B\plus{}C\plus{}D\minus{}A \equal{} 2,C\plus{}D\plus{}A\minus{}B \equal{} 2,D\plus{}A\plus{}B\minus{}C \equal{} 2,A\plus{}B\plus{}C\minus{}D \equal{} p^{2}\plus{}1$. Also using $ A \equal{} B \equal{} C$ this becomes $ A\plus{}D \equal{} 2,3A\minus{}D \equal{} p^{2}\plus{}1$ which has solutions $ A \equal{}\frac{p^{2}\plus{}3}{4},D \equal{}\frac{5\minus{}p^{2}}{4}$. But such $ p$ don't exist.

Now let $ (s\minus{}c,s\minus{}d) \equal{} (p,p)$. Then $ A \equal{} B$ and $ C \equal{} D$ and our system is $ 2C \equal{} 2,2A \equal{} p\plus{}1$, which gives $ (a,b,c,d) \equal{} (1,1,p,p)$.

So the only such quadrilateral is $ (1,1,p,p)$ and permutations.[/hide]

EDIT: Sorry when I started to write down my solution, there were no others, but now I see two already posted. :blush:


Let $a, b, c$ three nonnegative real numbers. Prove the inequality

                        \[ \sum_{cyc}\frac{a\sqrt{a^{2}+3bc}}{b+c}\geq a+b+c. \]

I think that

$\sum\frac{2a^3+6abc}{a^2+b^2+c^2+5bc}\geq a+b+c$ is true and stronger than original one .

Notice that $\frac{a\sqrt{a^2+3bc}}{b+c}=\frac{a^3+3abc}{(b+c)\sqrt{a^2+3bc}}\geq\frac{2a^3+6abc}{a^2+b^2+c^2+5bc}$ by am-gm .

Very nice your hint, Shyong. I think it works for Cezar's inequality. :lol:


search
geometry
geometric transformation
reflection
LaTeX
\/closed

I think the search functionality on this website could be greatly improved. For instance, I still cannot figure out how to search for a multiple word string like "big O notation". Also, the default for "Display results as" should definitely be posts. It extremely unhelpful to return a 25 page thread that we then have to mine for our term.

[quote="FibonacciFan"]I think the search functionality on this website could be greatly improved.[/quote]

Indeed Valentin (says he) is working on it. ;)

[quote="FibonacciFan"]Also, the default for "Display results as" should definitely be posts. It extremely unhelpful to return a 25 page thread that we then have to mine for our term.[/quote]

Display by topics is perfectly okay, at least as long as topic names reflect the contents of a topic. If they don't, this is a mistake on the side of the poster (I, as a mod, tend to edit topic titles in such cases).
Display by posts is pretty useless, at least in the way it is organized now, showing only the very beginning of the post, with broken LaTeX, and given that most posts are understandable only in the context of the thread, one has to open the thread so or so. It would be more useful if it would be possible to view only a sub-interval of the set of posts in a topic and then, via special buttons, extend this interval in any direction (i. e. view post #3, find out that one needs post #2 to understand what is said there, click to see post #2 as well, then wonder about possible replies and click on another button to see post #4 too).

Mathematical topics seldomly grow 25 pages long by nature. Usually, long topics are either problem marathons (pretty stupid idea, in my opinion, since this turns reading them a posteriori into a torture), or discussions with a lot of "junk" (not spam, but just stuff that can be removed after a certain time - e. g., posts of the kind "can you post your solution?" or posts reporting typos in other posts). Or the usual G&FF kind of discussion.

  Darij


calculus
integration
limit
trigonometry
logarithms
calculus computations

Evaluate

\[\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin ^ 2 nx}{\sin x}dx-\sum_{k=1}^n \frac{1}{k}\right)\]

Let $I(n) = \int_0^\pi \frac{\sin^2(nx)}{\sin x} \ dx$, after looking at a few values of $n$, I saw that

$I(n) = 2\sum_{k=1}^n \frac{1}{2k-1}$

And thus the original problem can be written as

$\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{1}{2k^2-k}\right)$

I think the above sum is equal to $2\ln 2$, but I can't prove it.

Yes, your answer is correct, Jimmy. :)

Let $J(n)=\sum_{k=1}^n \frac{1}{k}$, we are to find $2J(n)=....$


Parallel lines are drawn at intervals d on a table. A needle of length 1(< d) is thrown at
random on the table. What is the probability that the needle will touch one of the parallels?

This is [url=http://en.wikipedia.org/wiki/Buffon%27s_needle#Lazzarini.27s_estimate]Buffon's needle experiment[/url].


Prove that

(a) $ \displaystyle\sum_{k\equal{}0}^{l}\dbinom{n}{k}\dbinom{m}{l\minus{}k}\equal{}\dbinom{n\plus{}m}{l}$

(b) $ \displaystyle\sum_{k\equal{}0}^{n}\dbinom{n}{k}^2\equal{}\dbinom{2n}{n}$

[hide="a"] $ (x \plus{} 1)^n (x \plus{} 1)^m \equal{} (x \plus{} 1)^{n\plus{}m} \Leftrightarrow$
$ \sum_{l\equal{}0}^{n\plus{}m} \left( \sum_{k\equal{}0}^{l} {n \choose k} {m \choose l\minus{}k} \right) x^l \equal{} \sum_{l\equal{}0}^{n\plus{}m} {n \plus{} m \choose l} x^l \Leftrightarrow$
$ \sum_{k\equal{}0}^{l} {n \choose k} {m \choose l\minus{}k} \equal{} {n \plus{} m \choose l}$

This is known as [i]Vandermonde's Identity.[/i]  There is also a combinatorial proof.  Can you find it? [/hide]
[hide="b"] Let $ n \equal{} m$ in the above. [/hide]

hey there.

i have to do exactly the same proof, however, i think i have to show before, that i can use this "multiplication" rule for the series:
\[ \left(\sum_{i\equal{}0}^{n} a_i x^i\right) \left(\sum_{i\equal{}0}^m b_i x^i\right)
 \equal{} \sum_{i\equal{}0}^{m\plus{}n} \left(\sum_{j\equal{}0}^i a_j b_{i\minus{}j}\right) x^i. \]
holds. How can I accomplish that? I kinda got started on it, but dont know for sure.

\[ \left(\sum_{i\equal{}0}^{n} a_i x^i\right) \left(\sum_{i\equal{}0}^m b_i x^i\right)
 \equal{} \left(\sum_{i\equal{}0}^{n} a_i x^i\right) \left(\sum_{i\equal{}n}^{m\plus{}n} b_{n\minus{}i} x^{n\minus{}i}\right)
 \equal{} \left(\sum_{i\equal{}0}^{n} a_i x^i\right) x^n \left(\sum_{i\equal{}n}^{m\plus{}n} b_{n\minus{}i} x^{\minus{}i}\right)
 \equal{}  \]

I think that is the way to go, as the [\ x^{-i} \] and the  [\ x^i \] will cancel, however i am not sure about how to take the sums together...

thanks a lot for your help,
hmer

[quote="hmer"]i think i have to show before, that i can use this "multiplication" rule for the series:
\[ \left(\sum_{i \equal{} 0}^{n} a_i x^i\right) \left(\sum_{i \equal{} 0}^m b_i x^i\right) \equal{} \sum_{i \equal{} 0}^{m \plus{} n} \left(\sum_{j \equal{} 0}^i a_j b_{i \minus{} j}\right) x^i.
\][/quote]

This can be taken as the definition of polynomial multiplication; you then have to prove that polynomial multiplication is commutative, associative, and distributive, which is a pain.

Consider, then, the following argument:  the coefficient of $ x^i$ in the product must be taken from the product of the coefficient of some term $ x^j$ in the first factor and the coefficient some other term $ x^{i\minus{}j}$ in the second factor; in fact, the coefficient of $ x^i$ is precisely the sum of all such products, since it is precisely these products that contribute to the coefficient of $ x^i$.

Alternately, you can prove the result using a double induction on $ n$ and $ m$, which is tedious.

Combinatorial proof in this particular case still seems to be most direct.

[hide="a"]
Say you have two disjoint sets, $ A$ (with m elements) and $ B$ (with n elements). To choose $ l$ elements from the union of the sets, there are $ \binom{m\plus{}n}{l}$ ways. However, you can choose 0 things from set A and $ l$ from set B, 1 from set A and $ l\minus{}1$ from set B etc. Which is the LHS.
[/hide]

[quote="Otávio_Menezes"]Prove that

(a) $ \displaystyle\sum_{k \equal{} 0}^{l}\dbinom{n}{k}\dbinom{m}{l \minus{} k} \equal{} \dbinom{n \plus{} m}{l}$

[/quote]
This [url=http://www.mathlinks.ro/viewtopic.php?t=171763]link[/url] is dedicated to all people,who is interested in this kind of problems.Good luck!


linear algebra
matrix
induction
combinatorics unsolved
combinatorics

[b]Problem 1. [/b]In the cells of square table are written the numbers $1$, $0$ or $-1$ so that in every line there is exactly one  $1$, amd exactly one $-1$. Each turn we change the places of two columns or two rows. Is it possible, from any such table, after finite number of turns to obtain its opposite table (two tables are opposite if the sum of the numbers written in any two corresponding squares is zero)?

[i] Emil Kolev[/i]

I think it is possible.

Let $S_n$ be the Symetric group, then, the table with $+1$, determines a permutation $\pi$, and the table with $-1$ determiens a permutation $\sigma$. Since changing lines is equivalnet to do transpositions, and transpositions generate $S_n$, we have that we can acomplish it if there exist permutations $a$, and $b$ such that:

$a\pi b=\sigma$ and $a\sigma b=\pi$

but this happens if and only if:

$a(\pi\sigma^{-1})^{-1}a^{-1}=(\pi\sigma^{-1})$

and

$b^{-1}(\pi^{-1}\sigma)^{-1}b=(\pi^{-1}\sigma)$

But since every permutation is in the same conjugacy class as its inverse (since they have the same cycle lenghts) we can find such $a$ and $b$.

Yes, I believe it's always possible to reach the opposite ($n\times n$) table.

If $A$ is our initial matrix, basically, what we want is to find two permutation matrices $P,Q$ such that $PAQ+A=0$.

First permute the rows so as to bring all $1$’s on the main diagonal. The $-1$’s lie on the squares $(i,\tau(i))$, where $\tau\in S_n$ is a permutation of $\overline{1,n}$. Since we can use induction on the number of cycles in this permutation, we may as well assume it’s an $n$-cycle $(\sigma(1),\ldots,\sigma(n)),\ \sigma\in S_n$. Let $U$ be the matrix obtained from the identity matrix by permuting its columns according to $\sigma^{-1}$. Then $UAU^t$ ($A$ is not the old matrix $A$, but the one obtained after performing the operations above) will have $1$ on positions $(i,i)$, and $-1$ on positions $(i,i+1)$ (all indices are modulo $n$). This means that we amy as wel assume that our matrix $A$ had this form to begin with.

Now let $\tau_1,\tau_2\in S_n$ be the permutations $i\mapsto n-i$ and $i\mapsto n+1-i$ respectively (again, everything is modulo $n$), and let $P,Q$ be the permutation matrices which have $1$ on positions $(i,\tau_1(i))$ (for $P$) and $(\tau_2(j),j)$ (for $Q$). Then $PAQ$ will be $-A$, as desired.

Edit:

Oops :)


geometry
perimeter
inradius
modular arithmetic
Diophantine Equations

Prove that there exist infinitely many positive integers $n$ such that $p=nr$, where $p$ and $r$ are respectively the semi-perimeter and the inradius of a triangle with integer side lengths.

[quote="pohoatza"]From $S = pr = \frac{p^{2}}{n}$ and $S=\sqrt{p(p-a)(p-b)(p-c)}$, we have that $p^{3}= n^{2}(p-a)(p-b)(p-c)$.

And now by denoting $p-a=x$, $p-b=y$, $p-c=z$, the above equality turns out into $(x+y+z)^{3}= n^{2}xyz$.

Now the problems turns out to show that the above equation has a solution in natural numbers for infinitely many $n$.

For that assume $z=k(x+y)$, $\forall k \in \mathbb{Z_{+}}$. Thus the equation becomes $(k+1)^{3}(x+y)^{2}=kn^{2}xy$, and now by setting $n=3(k+1)$, we have that $(k+1)(x+y)^{2}=9kxy$.

Denote $t=\frac{x}{y}$. Thus the above equation has solutions in positive integers if and only if $(k+1)(t+1)^{2}=9kt$ has a rational solution, which is equivalent in proving that its discriminant $D=k(5k-4)$ is a perfect square.

Thus the problem turns out into proving that the equation $5k^{2}-4k=t^{2}$ has infinitely many integer solutions, which looks like Pell.[/quote]

The equation $5k^2 - 4k = t^2$ indeed looks like Pell, smells like Pell, and reduces to Pell, since it may be written as $(5k-2)^2 - 5t^2 = 4$, or $A^2 - 5B^2 = 4$ (of course, out of its infinitely many solutions, one needs to ascertain that infinitely many will have $A \equiv -2 \pmod{5}$, for example the primitive solution $(A_0,B_0) = (3,1)$).


$ a_{n}\equal{}a_{1}\plus{}(n\minus{}1)d$.  We have $ 91\equal{}1\plus{}(n\minus{}1)d\Rightarrow 90\equal{}(n\minus{}1)d$.  So I guess the answer would be the number of integer solutions to this equation.

[quote="JRav"]$ a_{n} \equal{} a_{1} \plus{} (n \minus{} 1)d$.  We have $ 91 \equal{} 1 \plus{} (n \minus{} 1)d\Rightarrow 90 \equal{} (n \minus{} 1)d$.  So I guess the answer would be the number of integer solutions to this equation.[/quote]

Factor 90! It factors into $ 2 \cdot 3^2 \cdot 5$, so we immediately see that there are $ 2 \cdot 3 \cdot 2 \equal{} 12$ factors. Thus, there are $ \boxed{12}$ such arithmetic sequences.

[quote="alkjash"]Wait..
Am I missing something?
I'm pretty sure that 90! does not factor as $ 2*3^2*5$[/quote]

I mean factor 90 (exclamation point).

LOL.

[quote="lotsofmath"]DonkeyKingKong, did you just post every MATHCOUNTS 2008 sprint problem???????????[/quote]

im pretty sure he did just post every sprint round question. what a spammer.

Eat my spam posters!!! :spam:  :spam:  :spam:

I could post this in every single one of his topics but that would be spamming.

Oh, ok.
I was pretty sure that 90! wasn't that small.  :D


A train normally travels at a uniform speed of $ 72 km/h$ on a long stretch of straight, level track.  On a particular day, the train must make a $ 2.0\minus{}$min stop at a station along the track.  If the train decelerates at a uniform rate of $ 1.0 m/s^2$ and, after the stop, accelerates at a rate of $ 0.50 m/s^2,$ how much time is lost because of stopping at the station?

I seems to be getting a different answer from the one in my book.  The book's answer is $ 180 s$ while mine is $ 420 s \equal{} 7 min.$

[hide="My own solution"] $ t \equal{} (v \minus{} v_0)/a \equal{} (0 \minus{} 72000/360 m/s)/(\minus{}1 m/s^2) \equal{} 72000/360 s \equal{} 200 s$ (actual time used before stopping)
$ x \equal{} 200 m/s * 200 s \minus{} \frac12 (m/s^2) (200 s)^2 \equal{} 200^2/2 m$
$ t \equal{} d/v \equal{} 0.5*200^2/200 s \equal{} 100 s$ (time used to cover the same distance if there's no deceleration, i.e., stopping)
$ 200 s \minus{} 100 s \equal{} 100 s$ (time wasted before stopping)

$ t \equal{} (v \minus{} v_0)/a \equal{} (200 m/s \minus{} 0)/0.5 m/s^2 \equal{} 400 s$
$ x \equal{} 1/2 at^2 \equal{} 1/4 (2*200 m)^2 \equal{} 200^2 m$
$ t \equal{} d/v \equal{} 200^2/200 m \equal{} 200 m$
$ 400 s \minus{} 200 s \equal{} 200 s$ (time wasted after stopping and beginning accelerating)

So the total time wasted is $ 100 s \plus{} 2 min \plus{} 200 s \equal{} 420 s \equal{} 7 min.$

[/hide]

Hey mathwizarddude! I think $ 72 km/h\equal{}\frac{72000}{3600}m/s\equal{}20m/s$....isn't it?? :maybe:
So the first time should be $ 20s$.

[quote="mathwizarddude"]$ t \equal{} (v \minus{} v_0)/a \equal{} (0 \minus{} 72000/360 m/s)/(\minus{}1 m/s^2) \equal{} 72000/360 s \equal{} \boxed{200s}$(actual time used before stopping)[/quote]
I think here is a mistake.
 :)

[color=darkblue]EDIT: Sunkern_sunflora already posted it - sorry ;)[/color]

Well, everything is fine, except the fact you've made a mistake in the calculation: $ 72 \frac {km}{h} \equal{} 20{m}{s}$, not $ 200$ as you have used ($ 200\frac {m}{s}$... whoa-it's not an airplane :) ). Correcting that, you get the stopping time of $ 20s$, accelerating time of $ 40s$, which adds up to $ 180s$ counting in the stopping.

But naturally, as you already pointed out, we have to subtract the time which the train would need to cover this distance without changing it's speed-and that's $ 30s$, so we get $ 150s$. And I'm pretty sure that's the correct answer.

I obtained similar results. Thanks hsiljak!
 :)

Yeah, thanks, how could I be so stupid as to used $ 1 h \equal{} 360 s$ when it should be $ 3600 s$ instead???  Is there any way to deal with such silly errors?  Thanks.

Don't take pressure for this....silly mistakes are inevitable.....everyone faces them, I mean, almost!
What we need to do is to keep cool and verify each n' every statements cautiously and throughly. I think it's the best way to stop silly mistakes. :D

Well, if I may be so bold and give an advice: Whenever doing a physics problem, make the model in your head, try to estimate some parameters without calculation (closely related: keep in mind some logical and real boundaries for those parameters, i.e.  $ 2 \frac{m}{s}$ is too slow for a train, $ 200 \frac{m}{s}$ is too fast). If the results obtained by calculations don't fit in those logical bounds, go back.


As a prank, Tim decides to steal Nathan's fork at dinner, but so he doesn't get caught, he convinces other people to do it for him. On Monday, he convinces Joe to do it. On Tuesday, he could get either Ambie or John to do it. On Wednesday, he can't convince any of those people to do it again, but there are five other people he could convince to do it. On Thursday, none of those five will do it, nor the first three, but there are four others who are willing. Finally, on Friday, Tim does it himself. How many different combinations of people could Tim involve in the prank?

Monday: 1 person
Tuesday: 2 people
Wednesday: 6 OR 5 people
Thursday: 4 people
Friday: 1 person

Depending how the question is interpreted, we have either 2*6*4 or 2*5*4, for a total of either 48 or 40.

[quote="GameBot"]How many different combinations of people could Tim involve in the prank?[/quote]
13 people could be involved in the prank 1+2+5+4+1 = 13.
Combinations of 13 people in groups of 1,2,3...13 = 8191

Number of people Tim could select to prank Nathan per day:

Monday:1
Tuesday:2
Wednesday:5
Thursday:4
Friday:1

So the answer is 2*5*4, or 40.

[quote=Williamz1483]some of you should hide your answers but there posts are really old so...[/quote]

ironic.

Also, poor Nathan :(  :noo: what did he ever do?

@above even more ironic .__.

nathan probably deserved it


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

a) Let $A\in\mathcal M_{n}(\mathbb Z)$ be a skew-symmetric matrix. Prove that $\det(A)$ is square of an integer.
b) Suppose $A=[a_{ij}]$ be a matrix in $\mathcal M_{n}(\mathbb R)$ that for each $i,j$, $a_{ij}=-a_{ji}$. Prove that $\det(A)=p(a_{ij})^{2}$, which $p$ is a polynomial of $a_{ij}$.

[hide="Remark"]We can construct a polynomial called Pffafian polynomial that $\det(A)=(\text{Pff}(A))^{2}$.[/hide]

I have tracked down the first part [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=matrix+antisymmetric&t=25572]here[/url].

In the link above, Kent has proved that it works for any field. Hence, if one considers the field $\mathbb{K}= \mathbb{R}(X_{i,j})_{1 \leq i,j \leq n}$ and the matrix $M \in M_{n}(\mathbb{K})$ defined by
$M_{i,j}=X_{i,j}$ for $i<j$ and
$M_{i,j}=-X_{j,i}$ for $i>j$
one knows there exists $F \in \mathbb{R}(X_{i,j})_{1 \leq i,j \leq n}$ such that
$\det(M) = F(X_{i,j})^{2}\in \mathbb{R}[X_{i,j}]_{1 \leq i,j \leq n}$
Hence, $F$ is actually in $\mathbb{R}[X_{i,j}]_{1 \leq i,j \leq n}$.

even it is wellknown that skew symmetric matrix can be represented $BJB^{T}$ when $J$ is block matrix which each block is $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ so the point the $det(A)$ is square of polynomial is easily deduced.


topology
advanced fields
advanced fields unsolved

Let $ M$ and $ N$ be two n-dimensional simplicial complexes.    Determine the first $ n$ simplicial homology group $ H_1( M \# N., \mathbb{Z})$

[quote="LkNsngth"]From the way your question is phrased, I'm not sure if you want help calculating $ H_1(M\#N)$ or $ H_k(M\#N)$ for $ 1 \leq k \leq n$. If its the latter, use the Mayer-Vietoris exact sequence. Define $ X = M\setminus B^n$ and $ Y = N\setminus B^n$ and let $ X \cup Y = M\#N$. Then $ X\cap Y = S^n$. Using the Mayer-Vietoris exact sequence, we get the following long exact sequence of singular homology:

$ H_{n + 1}(M\#N)\rightarrow H_n(S^n)\rightarrow H_n(X)\oplus H_n(Y)\rightarrow H_n(M\#N)\rightarrow H_{n - 1}(S^n)\rightarrow H_{n - 1}(X)\oplus H_{n - 1}(Y)\rightarrow H_{n - 1}(M\#N)\rightarrow H_{n - 2}(S^n)...$
Now, we know that because $ M\#N$ is an $ n$ manifold, all homology groups in dimension greater than $ n$ vanishes, so the first row becomes a short exact sequence which gives an isomorphism $ (H_n(X)\oplus H_n(Y))/H_n(S^n)\cong H_n(M\#N)$
In lower dimensions, we get two groups separated by trivial groups, so we get isomorphisms  $ H_k(X)\oplus H_k(Y) \cong H_k(M\#N)$

This reduces our problem to just calculating $ H_k(X)$ and $ H_k(Y)$ for various $ 0 \leq k \leq n$

We will now use excision. Let A be a slightly bigger ball which contains the $ B^n$ we removed from our manifold $ M$ (or $ N$). We want the homotopy type of $ A \setminus B^n$ to be $ S^n$ Then the excision theorem gives an isomorphism:
$ H_k(M \setminus B^n,A \setminus B^n)\cong H_k(M,A)\cong H_k(M)$ if A is suitably chosen (ie, its small enough such that it doesn't change the structure of the manifold).

Now apply the long exact sequence of relative homology:
$ H_{k}(M \setminus B^n)\rightarrow H_{k}(M \setminus B^n) \rightarrow H_{k}(M \setminus B^n,A \setminus B^n) \rightarrow ...$ This long exact sequence collapses into the following short exact sequence when $ k = n$: $ 0\rightarrow H_n(A \setminus B^n)\rightarrow H_n(M \setminus B^n)\rightarrow H_n(M \setminus B^n,M \setminus A^n)\cong H_n(M) \rightarrow 0$.
In dimensions less than $ n$, we get isomorphisms $ H_k(M \setminus B^n)\cong H_k(M)$
This concludes our calculation.[/quote]

I am sorry I was asking just for the first homology group.     I haven't seen the Mayer-Viertoris exact sequence and I haven't even seen yet the singular homology that's why I was asking about simplicial homology.

Thanks anyways.

do you know Hurewicz's theorem (it says that $ H_1(X)\cong \pi_1(X)/[\pi_1(X),\pi_1(X)]$ where the group on the right hand side if the isomorphism is the abelianization of the fundamental group? I think there might be a similar way of computing $ H_1(M\#N)$ using the Seifert-Van Kampen theorem for the fundamental group and then abelianizing it. (In fact, the Seifert-Van Kampen theorem is very similar to the Mayer-Vietoris sequence)

LkNsngth's solution is not quite 100% correct, because the question does not state that $ M$ and $ N$ are closed manifolds.  Consider the case when $ M$ and $ N$ are closed 2-disks, in which case the connected sum is a finite closed cylinder (aka a closed annulus).  (Other minor quibbles:  $ S^n$ shoud be $ S^{n\minus{}1}$, and replacing $ n$ by a general $ k$ is a bit confusing.)

Also, from what LittleFermat says, it sounds as if the question was intended to be solved with "bare hands."  That is, thinking of $ M$ and $ N$ as simplicial complexes, form the connected sum by removing one $ n$-face from both $ M$ and $ N$ and then identifying the boundaries of the removed $ n$-faces.  It's easy to see that this procedure has no effect on the boundary operator from $ k$-chains to $ k\minus{}1$ chains as long as $ k<n\minus{}1$.  So for $ n\geq3$, it's clear that
$ H_1(M\#N)\equal{}H_1(M)\oplus H_1(N)$.  Also, the $ n\equal{}1$ case is boring, so the only interesting case is $ n\equal{}2$.

The general $ n\equal{}2$ case is annoying (because it necessarily devolves into cases), but if you restrict yourself to the case of, say, orientable closed 2-surfaces, then a "bare hands" proof is reasonable.  You have to think about how the connect sum process affects the cycles and boundaries of $ M$ and $ N$.  Writing such a bare hands proof would amount to a concrete explanation of what's going on in the Mayer-Vietoris and excision argument described by LkNsngth.