TITLE: Prove that $\sqrt{5n+2}$ is irrational
QUESTION [9 upvotes]: I'm trying to follow this answer to prove that $\sqrt{5n+2}$ is irrational. So far I understand that the whole proof relies on being able to prove that $(5n+2)|x^2 \implies (5n+2)|x$ (which is why $\sqrt{4}$ doesn't fit, but $\sqrt{7}$ etc. does), this is where I got stuck. Maybe I'm overcomplicating it, so if you have a simpler approach, I'd like to know about it. :)
A related problem I'm trying to wrap my head around is: Prove that $\frac{5n+7}{3n+4}$ is irreducible, i.e. $(5n+7)\wedge(3n+4) = 1$.

REPLY [3 votes]: Since
$$
\{0,1,2,3,4\}\overset{x^2}{\longrightarrow}\{0,1,4\}\pmod{5}\tag{1}
$$
and
$$
2\not\in\{0,1,4\}\pmod{5}\tag{2}
$$
we know that $2$ is not a square mod $5$. That means that
$$
5n+2=x^2\tag{3}
$$
has no solutions with $n,x\in\mathbb{Z}$.
If $\sqrt{5n+2}\in\mathbb{Q}$, then there is some $x\in\mathbb{Q}$ so that
$$
x^2-(5n+2)=0\tag{4}
$$
which implies that $x$ is a rational algebraic integer, therefore, by this answer, it must be an integer. However, this contradicts that $(3)$ has no integer solutions.
Therefore,
$$
\sqrt{5n+2}\not\in\mathbb{Q}\tag{5}
$$

Since
$$
3(5n+7)-5(3n+4)=1\tag{6}
$$
we know that $5n+7$ and $3n+4$ have no common factors.