show that if $\displaystyle 19a^{2} \cong b^{2} (mod\ 7)$ then $\displaystyle 19a^{2} \cong b^{2} (mod\ 7^{2})$
First note that $\displaystyle 19\equiv5\!\!\!\pmod7$. So we want to know when a square can be congruent to five times a square (mod 7).
The only squares (mod 7) are 0, 1, 2 and 4. And the only way for one of these to be congruent to five times another one is if they are both 0. Thus a and b must both be congruent to 0 (mod 7), and so $\displaystyle a^2$ and $\displaystyle b^2$ must both be congruent to 0 (mod $\displaystyle 7^2$).