A number theory problem by Fahim Shahriar Shakkhor

Case 1

$n$ is even. Then the LHS is a sum of perfect squares, and perfect squares are non-negative. Equality holds if and only if $\displaystyle\begin{cases}x=0\\x+2=0\\2-x=0\end{cases}$

But this cannot be true for a fixed $x$ . Therefore, no solutions $x$ exist when $n$ is even.

Case 2

$n$ is odd. $x^n+(x+2)^n=(x-2)^n$

$x, x+2, x-2$ are integers and $n$ is a positive integer. We can apply Fermat's Last Theorem . There are no integer solutions $x$ when $n>2$ , hence the only odd possibility of $n$ is $1$ . After checking the case $n=1$ , it's clear that $x=-4$ , which is an integer.

Hence the only possibility is $n=1$ . The answer is $\boxed{1}$ .