If $x$ and $y$ are from $[2, 100]$ show there exist $n$ so $x^{2^n}+y^{2^n}$ is composite. If $x$ and $y$ are form $[2, 100]$ show there exist $n$ so $x^{2^n}+y^{2^n}$ is composite.
This is old contest problem and I can't find solution. 
I tried to show that for some $n$ this sum will be divisible by 101, but didn't succeed.
Also tried some more ides, but also failure.
Also we can assume that $(x,y)=1$.
Can someone help? Any hint or idea what to try?
 A: Here's one way to do it, though maybe it's not the best way. It relies on the Fermat number $257 = 2^{8}+1$ being prime.
By Fermat's little theorem, since $257$ is prime, we have 
$$
   2^{256} \equiv 3^{256} \equiv 4^{256} \equiv \dots \equiv 100^{256} \equiv 1 \pmod{257}.
$$
So, for any $x, y \in [2,100]$, $x^{256} - y^{256}$ is definitely divisible by $257$.
But we can factor $x^{256} - y^{256}$ as
$$
   (x-y)(x + y)(x^2 + y^2)(x^4 +y^4)(x^8+y^8) \dotsm (x^{128} + y^{128}).
$$
So one of these factors must be divisible by $257$.  It cannot be $x-y$, because $x-y \in [-98, 98]$, so it must be one of those other factors.
This immediately tells us that one of the other factors is composite, unless one of the other factors is equal to $257$ exactly. However:


*

*We cannot have $x+y = 257$, because $x+y \le 200$.

*We cannot have $x^2 + y^2 = 257$, because the only integer solution to this is $(x,y) = (\pm1, \pm16)$ and $(x,y) = (\pm16, \pm1)$. One of $x$ or $y$ would have to be $1$, which is impossible.

*We cannot have $x^{2^n} + y^{2^n} = 257$  for any $n>1$, because this would correspond to $x'^2 + y'^2 = 257$ for $x'= x^{2^{n-1}}$ and $y'= y^{2^{n-1}}$, and we return to the previous case.


So one of the factors $x+y, x^2 + y^2, \dots, x^{128} + y^{128}$ is divisible by $257$ but not equal to it, and therefore it must be composite.
