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?
 A: Hint:
If $3\not\mid x,$ then $x^2\equiv1\pmod3.$
Apply this to $x$ and $y,$ and conclude.  
Hope this helps.
A: Factoring, is not the best way.  But for fun let us use factoring. 
If neither $x$ nor $y$ is divisible by $3$, then $x\equiv y\pmod{3}$ or $x\equiv -y\pmod{3}$. In the first case, $3$ divides $x-y$. In the second, $3$ divides $x+y$. So in either case $3$ divides $(x-y)(x+y)$.
A: If $3 \nmid x$, then $x = 3k + 1$ or $3k + 2$. In the former case, $x^2 = (3k + 1)^2 = 9k^2 + 6k + 1$ which can be restated as $3j + 1$, while in the latter case, $x^2 = (3k + 2)^2 = 9k^2 + 12k + 4$, which can also be restated as $3j + 1$. In the notation of congruences, if $3 \nmid x$, then $x^2 \equiv 1 \pmod 3$.
Let's say $x^2 = 3j + 1$ and $y^2 = 3h + 1$. Then $$x^2 - y^2 = (3j + 1) - (3h + 1) = 3j - 3h = 3(j - h).$$ With congruences, you can just do $x^2 - y^2 \equiv 1 - 1 = 0 \pmod 3$.
A: 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.

