if $19a^2 \equiv b^2 \pmod 7$ then $19a^2 \equiv b^2 \pmod {7^2}$

Question

I am stuck with this problem. All what I can tell is that $19a^2 \equiv 5a^2 \equiv b^2 \pmod 7$ and $5$ is not a quadratic residue$\pmod 7$. Any hints please,,

Answer 1

Suppose that $a$ is nonzero on $\Bbb Z_7$, so it has an inverse $a^{-1}$. Then $\overline 5 \equiv {(a^{-1})}^2b^2 \equiv (a^{-1}b)^2 \equiv c^2$. But $5$ is not a quadratic residue modulo 7, so...