Can the irrationality of the square root of 2 be proved by using Dirichlet's theorem on primes in an arithmetic progression? The title says it all.
I intend to answer the question myself, in the affirmative.
(I would have left the body blank, but the system requires me to post at least 30 characters.)
 A: Here is an alternative proof (this is ridiculous, but fun!). Suppose that $(a/b)^2 = 2$. This equation can be seen as an equation in $\mathbf F_p$, whenever $p \nmid b$. On the other hand, Gauss shows that for an odd prime $p$,
$$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/2}.$$
This is $-1$ if $p \equiv \pm 3 \mod 8$ and $+1$ if $p \equiv \pm 1 \mod 8$.
By Dirichlet's theorem on primes in arithmetic progressions, there exists an odd prime $p$ such that $\left(\frac{2}{p}\right) = -1$ and $p \nmid b$. But this contradicts the fact that $(a/b)^2 = 2$ in $\mathbf F_p$.
A: Yes.
Dirichlet's theorem states that every proper arithmetic progression contains infinitely many primes.
Theorem. The square root of $2$ is irrational.
Proof: Suppose that $a$ and $b$ are relatively-prime positive integers such that $(a/b)^2 = 2$.
Then, by inspection, $b > 1$, and so $b^2$ is composite. Since $a$ and $b$ are relatively prime, so are $a^2$ and $b^2$. By Dirichlet's theorem, there exists a positive integer $n$ and a prime $p$ such that
$a^2 + nb^2 = p$. Then, dividing through by $b^2$, we have $a^2/b^2 + n = p/b^2$.
Then, since $a^2/b^2 = (a/b)^2$, we have $(a/b)^2 + n = p/b^2$.
Then, since $(a/b)^2 = 2$, we have $2 + n = p/b^2$.
However, the left side is an integer, but the right side cannot be an integer, since it is the ratio of a prime to a composite number. This contradiction shows that the square root of $2$ must be irrational.
Q.E.D.
