Stuck again! By the way thanks again for all the help!

Prove or disprove:

For all positive integers a, b: if $\displaystyle a^2$ $\displaystyle \equiv$ $\displaystyle b^2$ (mod 9), then a $\displaystyle \equiv$ b (mod 9)

I haven't found a counterexample so it looks true and I have the proof for the converse. I've only gotten this far:

Given

$\displaystyle a^2$ $\displaystyle \equiv$ $\displaystyle b^2$ (mod 9)

then 9 | ($\displaystyle a^2$ - $\displaystyle b^2$) = (a - b)(a + b)

I know that if s | tu then s | t or s | u.

So 9 | (a - b) or 9 | (a + b) and we need to show that 9 | (a - b) but I'm not sure where to go from here.

Thanks