<s>[INST] Let x and y be integers, prove that if 3 doesn't divide x and 3 doesn't divide y then 3 divides $x^2 - y^2$ Let x and y be integers, prove that if $3 \nmid x$ and $3 \nmid y$ then 
$3 \mid (x^2 - y^2)$
Attemmpt:
The only thing i get out of this is that there is a difference of squares: $$(x^2 - y^2) = (x-y)(x+y) $$
Other than that i am  stuck. Perhaps a hint before a full blown solution?
 [/INST] Note that $3$ does not divide $x$ and $y$ means that $3$ does not divide $x^2$ and $y^2$.
Hence, $x^2$ and $y^2$ leave remainder 1 when divided by $3$. This is a simple rule of perfect square numbers.
Let $x^2=3q+1$ and $y^2=3p+1$.

$\therefore x^2-y^2=3q+1-3p+1=3(q-p)\implies 3$ is a factor of $x^2-y^2$.
Thus, the answer follows.

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