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.


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*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?

$1995$ I'd divisible by $3$ but not by $9$ hence $x , y $ 
 both simultaneously are not multiple of $3$.
$$ $$ let either of $x,y$ is multiple of $3$ and other is not multiple of $3$ then $x^2-y^2$ is not multiple of $3$ where as $ 1995$. Is multiple of  $3$ hence not possible.
