Exercise 3.2.17* ($19a^2 \equiv b^2 \pmod 7 \Rightarrow 19a^2 \equiv b^2 \pmod{7^2}$)

Proof. Suppose that $19a^2\equiv b^2(\mathit{mod}7)$ for some integers $a,b$. Since $19\equiv 5(\mathit{mod}7)$, we obtain $5a^2\equiv b^2(\mathit{mod}7)$.

Assume for contradiction that $7\nmid a$. Then $a$ has an inverse $\bar{a}$ modulo $7$, which satisfies

Therefore $5$ is a quadratic residue modulo $7$, thus

But $5\equiv 1(\mathit{mod}4)$, hence

This is a contradiction, which proves that $7\mid a$.

Since $7\mid a$, and $5a^2-b^2=7k$ for some integer $k$, $7\mid b^2$, thus $7\mid b$ because $7$ is prime. Therefore $7^2\mid a^2,7^2\mid b^2$, so

If $19a^2\equiv b^2(\mathit{mod}7)$, then $19a^2\equiv b^2(\mathit{mod}{7}^2)$. □