# Thread: Prove a congruence implication

1. ## Prove a congruence implication

show that if $19a^{2} \cong b^{2} (mod\ 7)$ then $19a^{2} \cong b^{2} (mod\ 7^{2})$

2. Originally Posted by mancillaj3
show that if $19a^{2} \equiv b^{2} \!\!\!\pmod 7$ then $19a^{2} \equiv b^{2} \!\!\!\pmod{7^{2}}$
First note that $19\equiv5\!\!\!\pmod7$. 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 $a^2$ and $b^2$ must both be congruent to 0 (mod $7^2$).