Here is my solution .We only need consider $ n>1$ The first we consider : $ a=\frac{x}{p},y=\frac{y}{q}$ where $ \gcd(x,p)=\gcd(y,q)=1$ and $ p,q>0$ The condition equivalent i)$ pq|xq+py$ From 1 we have $ q|p,p|q$ so $ p=q$ So we only need to consider when $ \gcd(x,y)=1$ Then condition equivalent : i)$ p|x+y$ $ p^n|x^n+y^n$ Case 1 $ n\equiv 0 (\mod 2).$ Because $ \gcd(x+y,x^n+y^n)\in\{1,2\}$ do so $ p=2$ But $ 2^n|x^n+y^n$ then $ 2|x,2|y$ illogic. Case 2 $ n\equiv 1 (\mod 2)$ Call p is a prime divisor of n . Then $ v_p (x^n+y^n)=v_{p}(x+y)+v_{p}(n)$ If we chose $ v_p(x+y)\geq v_{n}(n-1)$ then it satisfies conditon . So answer is $ n\equiv 1 (\mod 2)$

What is your function $ v$? Is it the power of $ p$ that divides $ n$? Why $ v_p (x^n+y^n)=v_{p}(x+y)+v_{p}(n)$?

To eay you only chose $ p^n|x+y$ This lemma posted many time on Mathlinks

TTsphn wrote: This lemma posted many time on Mathlinks Which lemma are you referring to?

Hi. I was just wondering where you got that problem from? It was in Croatia TST a week ago, but I'm sure it's from some older competition. I'd just like to know which. I've solved it (Wee!) so I'd like to know where it's from (I actually develop a sort of an emotional bond with the not-so-easy problems I solve on important competitions ). Thanks a lot!

Well, mornik, I took it from Croatia TST It's in the imo.org.yu website. Anyway, for the $ n$ odd case, I guess we can just choose $ a=-b$ not being integers. But what if we want positive $ a,b$? Can you help me?

Yes chose $ p^n|x+y$ and $ \gcd(x,y)=1$ There exist infinite such pair $ (x,y)$

Simply choose $ a = \frac {2^n + 1}{2}$ and $ b = \frac {2^n - 1}{2}$ and you got yourself a solution for odd $ n$.