The Proof is Trivial!

OK, I won't.
Is that problem 161? Looks like a mess, but I might take a punt, right now I'm having a go at some of the others that you just posted

To demonstrate that these are the only solutions with the given restrictions..

Here's my "progress post", updated as I solve the question

k=1 case: I used Mathematica to get a couple of hundred iterations, I just couldn't intuit the limit… then it was easy enough to show by induction that $b_n \equiv \dfrac{a_n^2}{n} > 2$ always, after getting a recurrence relation for $b_n$. Also shown previously that $b_n$ is strictly decreasing. Now I just need to show that 2 is indeed the limit.

Solution 307

Solution 310

\displaystyle \frac{1}{2} \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{m^2 n(3^n m)+ n^2 m (3^m n)}{3^{m+n} (n 3^m + m3^n)} \right)

It's actually really neat and, dare I say, 'straightforward' provided that you have the right tools. The hardest part is having the determination to persist (because everything seems to feel like a dead end in this problem - even if it's along the right lines!)

Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'-status suggests!

OK, I'll give it a go tomorrow.
I figured irrationality looked like a faff of writing words, to be honest the quadratic looks messy but is actually quite simple
I've spent a while on Problem 305.

Problem 308*

Find all pairs of integers (x,y) such that $\displaystyle 1+x^2y=x^2+2xy+2x+y$

Problem 314*

Find all positive integers n such that $n+2012$ divides $n^2+2012$ and $n+2013$ divides $n^2+2013$.

Problem 305*

Prove that the sequence defined by $y_0=1$, $\displaystyle y_{n+1}=\frac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right)$ for $n \ge 0$ consists only of integers.

[As a bonus, can you find any 'similar' sequences for which this is true?]

Awesome solutions bro, I did the same in both cases. Also, this line in a typo^

Problem 312*

Evaluate $\displaystyle \frac{4}{4^2-1}+\frac{4^2}{4^4-1}+\frac{4^4}{4^8-1}+\frac{4^8}{4^{16}-1}+\cdots$

Problem 312*

Evaluate $\displaystyle \frac{4}{4^2-1}+\frac{4^2}{4^4-1}+\frac{4^4}{4^8-1}+\frac{4^8}{4^{16}-1}+\cdots$

Solution 314

I'm posting this, because I think mine's one way of looking at it, but I'm concealing it because it's a nice one that I hope others solve as well.

Cheers, man. These were a good bunch, eh?

An ugly solution, but a solution nonetheless.

Problem 304*

Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely?

Solution 304

There exists no such integer k....

If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin.

I'll wait for others to post a solution to this one first, but I just wanted to point out something.

Did anybody else notice that $y_n = F_{2n+1}$, where $F_n$ denotes the nth Fibonacci number?
So, this sequence, essentially generates every other Fibonacci number.

Problem 305*

Prove that the sequence defined by $y_0=1$, $\displaystyle y_{n+1}=\frac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right)$ for $n \ge 0$ consists only of integers.

[As a bonus, can you find any 'similar' sequences for which this is true?]

If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin.

Problem 316*

Prove that
$\displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2}= \frac{\pi^4}{120}$
and find a general form for
$\displaystyle \sum_{x_1<x_2<...<x_n, x_1,x_2,...,x_n \in \mathbb{N}}\frac{1}{\prod_{k=1}^nx_k^2}$

Solution 316
...
And herein lies the crux move. Consider the co-efficient of $y^2$ in this expansion.

We have:

Co-efficient of $y^2 = \displaystyle \frac{1}{\pi^4} \left( \left(\frac{1}{2^2}+ \frac{1}{3^2}... \right)+ \frac{1}{2^2} \left(\frac{1}{3^2}+ \frac{1}{4^2}...\right) ... \right) = \frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2}$.

And voila, this is (almost) the very expression we had in the first line.

Notice that in the Maclaurin expansion of $\displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}}$, we had the co-efficient of $y^2 = \frac{1}{5!}$.

Equating the 2, we have:

$\displaystyle\frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{1}{5!}[br][br]\Rightarrow \displaystyle \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{\pi^4}{5!}[br][br]\Rightarrow \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2} = \frac{\pi^4}{5!}$

...

Problem 317**


Find

$\displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{2k+1}-\frac{2}{\pi(2k+1)^2}$

Solution 317 (I think?)

Does this not require absolute convergence? You've split up the summand, which you can't necessarily do without absolute convergence.

Hence my hesitation. In fact, I don't think it is.(?)