This is not a complete solution, but I guess this is the right way to do it: put $ a_n=cb_n$, then $ c^2(c+b_n^2)=x^m$ is pretty trivial to discuss when $ m$ is even, because obviously $ (b_n,c)=1$ and so it would follow that $ b_n^2+c$ is a square, thus greater than $ (b_n+1)^2$, from where we get an inequality which bounds $ b_n$ so much that the conclusion is trivial. The interesting case is when $ m$ is odd, so $ c$ and $ c+b_n^2$ should be $ m$-th powers, since they are relatively prime. Now, $ b_n$ is a polynomial in $ c$ and some manipulations show that $ b_n^2+c=(1+c)s$ with $ s$ relatively prime to $ c$. However, this is not easy to prove, at least I do not see a nice way that would avoid some horrible computations. Hope to see something nice. Note that once this is done, the rest is obvious, as $ c$ and $ c+1$ being $ m$-th powers does not happen too often. Don't be too bad with me for this...

harazi wrote: This is not a complete solution, but I guess this is the right way to do it: put $ a_n = cb_n$, then $ c^2(c + b_n^2) = x^m$ is pretty trivial to discuss when $ m$ is even, because obviously $ (b_n,c) = 1$ and so it would follow that $ b_n^2 + c$ is a square, thus greater than $ (b_n + 1)^2$, from where we get an inequality which bounds $ b_n$ so much that the conclusion is trivial. The interesting case is when $ m$ is odd, so $ c$ and $ c + b_n^2$ should be $ m$-th powers, since they are relatively prime. Now, $ b_n$ is a polynomial in $ c$ and some manipulations show that $ b_n^2 + c = (1 + c)s$ with $ s$ relatively prime to $ c$. However, this is not easy to prove, at least I do not see a nice way that would avoid some horrible computations. Hope to see something nice. Note that once this is done, the rest is obvious, as $ c$ and $ c + 1$ being $ m$-th powers does not happen too often. Don't be too bad with me for this... From $ c^2(b_n^2+c)=x^m$ and $ \gcd(c^2,b_n^2+c)=1$ so $ c^2=a^m,b_n^2+c=b^m$ So $ m\equiv 0(\mod 2)$

You're wrong. Meantime, I have found something that proves my assertion: put $ x_n = \frac {b_n - 1}{c(1 + c)}$, then computation (I hope this is not wrong, otherwise all that follows is stupid) gives $ x_{n + 1} = 1 + 2cx_n + c^2(1 + c)x_n^2 + x_n$. On the other hand, computation gives $ b_n^2 + c = (1 + c)(c^2(1 + c)x_n^2 + 2cx_n + 1)$. Now, the recurrence relation gives $ x_{n + 1} = 1 - x_n$ mod $ c + 1$ and since $ x_1 = 0$ then all $ x_j$ are $ 0$ or $ 1$ mod $ c + 1$ and so $ 2cx_n + 1$ is either 1 or $ - 1$ mod $ c + 1$, finishing the claim. Hope it's not completely wrong, anyway nice problem. [edit: cultural question] By the way, does anyone know relevant results on the equation $ x^2+y^m=z^m$ for odd $ m$?

Please can anyone solve last equation that harazi wrote Then for sure I can say that I will receive a medal. Please!

harazi wrote: [edit: cultural question] By the way, does anyone know relevant results on the equation $ x^2 + y^m = z^m$ for odd $ m$? I think it's still a conjecture: http://mathworld.wolfram.com/Fermat-CatalanConjecture.html Sorry delegat

Please can you have a look at http://www.mathlinks.ro/viewtopic.php?t=203664 This is equation that I can not solve. I would be really thankful if someone can finish this.

It is a little tricky but all one needs to observe is that $ b_n\equiv 1, - c,1, - c,\dots \pmod{(c + 1)^2}$ this doesn't require much effort.

Here's mine: Put a{n}=c(c*b{n}+1), then b{n+1}=b{n}+c+(c*b{n}+1)^2, b{1}=0, hence all integers. Then put b{n+1}-b{n}=t{n}, and we get also that all t{i} are integers satisfying reccurence t{n+2}=(c*c*t{n}*t{n+1}*(t{n}+t{n+1})+t{n+1}^2), and by easy induction we get all t{i} are divisible with c+1 and t{i}/(c+1) is of form c(c+1)K+1. Now since a{i}^2+c^3=c^2*t{i}, we now easily get that only solution is c=a^2-1.

BTW, I liked the official solution very very much!!

Can somebody send the official solution?

http://www.smm.org.mk/images/pdf/problems.pdf There you go...

thanks for the solution.

which is: ?

Albanian Eagle wrote: which is: ? This and all these other methods ,axioms and theorems of omerftekin that trivializes the IMO and BMO problem are just secret Albanian Eagle . Maybe all these things are the problems theirself But as harazi would say : These theorems are too big for us , the idiots

these axıoms methods theorems lemmas are very well known benefıtıcal arguments for bmo ımo.don't you know them

no why dont you show us.

Let $b_n = a_n^2 + c^3$ - we are interested in a $b$-s to be a perfect power. The pretty much key point is the recursion $b_{n+1} = a_{n+1}^2 + c^3 = (a_n^2 + a_n + c^3)^2 + c^3 = (b_n + a_n)^2 + c^3 = b_n^2 + 2a_nb_n + a_n^2 + c^3 = b_n(b_n + 2a_n + 1)$ as well as the guess that $b_n$ and $b_n + 2a_n + 1$ are coprime. To prove the latter, let $p$ be a common prime divisor - then $p$ must divide $2a_n + 1$ and so $2a_n \equiv -1 \pmod p$, leading to $0 \equiv 4b_n = (2a_n)^2 + 4c^3 \equiv 4c^3 + 1 \pmod p$, so $-2 \equiv 4a_n = 4a_{n-1}^2 + 4a_{n-1} + 4c^3 = (2a_{n-1}+1)^2 + 4c^3 - 1 \pmod p$, i.e. $p$ divides $2a_{n-1} + 1$ - so by induction it follows that $p$ must divide $2a_1 + 1 = 2c + 1$, but then $4c^3 \equiv - 1 \pmod p$ and $2c\equiv -1 \pmod p$ imply $c\equiv -1 \pmod p$ and $p\mid 1$, contradiction! Therefore if $b_{n+1}$ is an $m$-th power, then so is $b_n$ - and now by induction we deduce that $b_1 = c^3 + c^2$ must also be an $m$-th power. However, $b_1 = c^2(c+1)$, with both factors clearly being coprime and hence $m$-th powers - so either $m\geq 4$ is even or $c$ is a perfect square and we get two positive squares with difference $1$ - impossible. So the only remaining option is $m=2$ and $c+1$ being a perfect square - therefore all possible $c$ are $\ell^2 - 1$ for any integer $\ell \geq 2$.