<s>[INST] If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3 
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.

I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.


*

*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3} 
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.


*If $x$ and $y$ are divisible by $3$:


Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?
 [/INST] The question is a bit ambiguous. Anyway, you can indeed answer both interpretations.
If both $x$ and $y$ are divisible by $3$, then the left-hand side is divisible by $9$, but the right-hand side isn't (which is your proof).
Suppose $x$ is divisible by $3$. Then $-y^2\equiv0\pmod{3}$, because $1995\equiv0\pmod{3}$. Thus implies $y\equiv0\pmod3$ and we already have excluded the case.
Similarly if $y$ is divisible by $3$.
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