Proving irrationality of $\sqrt n$ if n is congruent to 3, 5, 6 mod 7

Question

I am unable to determine how to prove the following question.

Suppose that $n \in \Bbb N$. Prove that $\sqrt n$ is irrational if $n \equiv 3, 5 \ or \ 6 (mod 7)$

Thanks

Answer 1

Suppose it is rational, then

$$ \sqrt{n}=\frac{a}{b}\Longrightarrow n=\frac{a^2}{b^2}\Longrightarrow n\equiv a^2b^{-2}\pmod{7} $$

and $a^2b^{-2}$ is a quadratic residue modulo 7. Such residues are $\{(\pm 1)^2,(\pm 2)^2,(\pm 3)^2\}=\{1,2,4\}$. Notice that $b^{-2}$ is the inverse of $b^2$, so $a^2b^{-2}\in\{1,2,4\}$ [*].

Since $n\equiv a^2b^{-2}\pmod{7}$, if $n$ is rational, it cannot be equivalent to 3, 5 or 6.

* the inverse of a quadratic residue is a quadratic residue