<s>[INST] $x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares 
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.

Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Suppose $x^2+xy+y^2=a^2$ and $x^2-xy+y^2=b^2$. Then $2xy=a^2-b^2=(a+b)(a-b)$, so one of $x$ and $y$ is even, but then I am stuck.
 [/INST] This is not yet a full answer, but I need to get ready for work now and I'm not sure how to proceed anyway.
This statement clearly doesn't hold if $x$ or $y$ can be $0$, and negative numbers are easily flipped about, so assume that $x$ and $y$ are positive integers.
If $x$ and $y$ are both even, then dividing each by $2$ will yield another solution. Thus if there is any solution, there must be one for which $x$ or $y$ is odd, so we will assume, without loss of generality, that $y$ is odd.
Since $2xy=(a+b)(a-b)$, either $a+b$ or $a-b$ must be even. But $a+b = (a-b)+2b$, so in fact $a+b$ and $a-b$ are both even.
Thus $4\mid (a+b)(a-b)=2xy$, so $x$ or $y$ is even. Since $y$ is assumed to be odd, $x$ must be even.
Thus $a$ and $b$ are both odd.
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