This is my first question here after a lot of lurking. I hope it's found to be at an appropriate level for the site.
This question found its way to me - it supposedly originates as an extra credit problem assigned to an advanced high school student, but I have some graduate level education in math and haven't been able to find an answer, so I'm skeptical.
Let $x,y,z\in \mathbb{Z}$ satisfy $x+y+z=0$. Consider the expression $\frac{x^n+y^n+z^n}{2}$. For what $n\in\mathbb{Z}$ is it true that our expression is automatically the square of some integer, regardless of the particular values of $x,y,z$?
It's easy to see that $n=1$ is one solution, since the expression is then identically zero.
One can also discover with some playing around that $n=4$ works, because it turns out that $\frac{x^4+y^4+(-x-y)^4}{2}=(\frac{x^2+y^2+(-x-y)^2}{2})^2$ - that is, the expression at $n=4$ is identically the square of the expression at $n=2$.
Empirical searches with computer software suggest that $n=1$ and $n=4$ are in fact the only solutions possible. Can anybody here prove this?
As a simple preliminary step, one can deduce that $n$ must be of the form $n=2^k$ for some integer $k$, or else the denominator $2$ in the expression will never cancel out and yield an integer. But at that point, I'm stuck.
Take $x=2$, $y=z=-1$. if $n=2m$ with $m\in\Bbb{N}$, we have: $$\dfrac{x^n+y^n+z^n}2=\dfrac{2^{2m}+2}{2}=2^{2m-1}+1$$ Suppose now that $$2^{2m-1}+1=k^2$$ $$2^{2m-1}=(k+1)(k-1)$$ So, $k+1$ and $k-1$ are powers of $2$, so $k=3\implies m=2\implies n=4$. $n=4$ was verified by you and now we know it's the only even option for $n$.
Now, take $n=2m+1$. We obtain: $$\dfrac{x^n+y^n+z^n}2=\dfrac{2^{2m+1}-2}{2}=k^2$$ $$(2^m)^2-1=k^2$$ The only two squares that differ by $1$ are $0^2$ and $1^2$, which gives $k=0\implies m=0\implies n=1$, which is the other validated option.