October 20th
Today I learned a proof for a slightly weaker version of Cohn's irreducibility criterion from Murty . Quite surprisingly, there is pretty much no number theory involved; it's pretty much entirely bounding. Fix our polynomial $f(x)=a_nx^n+\cdots+a_0\in\ZZ[x].$ The bounding constant of interest is\[H=\max_k|a_k/a_n|.\]We're not going to state the exact statement until the end of the proof.
We begin by showing that all roots $\alpha$ of $f$ satisfy $|\alpha|\le H+1.$ Note that $|\alpha| \lt 1$ makes this immediate. Otherwise, $f(\alpha)=0$ implies\[\alpha^n=-\frac1{a_n}\left(a_{n-1}\alpha^{n-1}+\cdots+a_0\right),\]so\[|\alpha|^n=\left|\sum_{k=0}^{n-1}\frac{a_k}{a_n}\alpha^k\right|\le\sum_{k=0}^{n-1}H|\alpha|^k=H\cdot\frac{|\alpha|^n-1}{|\alpha|-1} \lt H\cdot\frac{|\alpha|^n}{|\alpha|-1}.\]Rearranging gives the desired $|\alpha| \lt H+1.$
To finish, assume (not for the sake of contradiction) $f(x)$ is irreducible, factored by $f(x)=g(x)h(x)$ with $g(x),h(x)\in\ZZ[x].$ We're going to introduce the exact criterion's condition later. We see that we may write $g(x)=\gamma\prod_k(x-\alpha_k)$ with $\gamma\ge1.$ Note roots $\alpha_k$ are also roots of $f(x),$ so for $x\ge H+2,$ we may bound\[|g(x)|=|\gamma|\prod_{k=1}^{\deg g}|x-\alpha_k|\ge\prod_{k=1}^{\deg g}(|x|-|\alpha_k|) \gt 1.\]The same holds for $h(x).$
We structured the proof this way so far in order to emphasize the lack of number theory we've done so far—everything has been bounding. Only now do we bring in any number theory. If we fix an integer $n\ge H+2,$ we note that $f(n)=g(n)h(n)$ makes a nontrivial factorization for $f(n)$ because $g(n),h(n) \gt 1$ from the above work. It follows that $f(n)$ must be composite. So we can say that $f(x)$ is irreducible if there exists an $n\ge H+2$ such that $f(n)$ is prime, and that's all.
It is conjectured that irreducible polynomials produce "a lot'' of primes (say, see the Bateman-Horn conjecture), even for these large (actually pretty small) values of $x,$ so what I like about this irreducibility criterion is that it actually does a pretty good job of deciding irreducibility. Namely, compute $H+2$ and then try a whole bunch of consecutive values of $x$ while testing for primes. For example, the polynomial $x^2-14x^2+9$ from two days ago has $H+2=16,$ and $20^4-14\cdot20^2+9=154409,$ which is prime. So irreducibility follows.
Notably, this is still not sharp because of polynomials like $x^2+x+2$ being irreducible but always even. But I'm still content that Cohn's criterion works for most sane polynomials. According to the Bunyakovsky conjecture, the condition we need to add is that $\gcd(f(n))=1$ over all $n,$ which is quickly checked by computing the greatest common divisor of the first few values.