下午从七分厂返回，和大 N 老师聊天提到数论，于是知道了这个证明.

Thm. For every fixed $m$, there are infinitely many prime numbers of the form $km+1$ where $k\in\mathbb{Z}$.

Lem 1. $n\gt1$, $p$ is prime, if for some $x\neq\pm1$, $p\mid\Phi_n(x)$, then either $p\mid n$, or $p\equiv_n1$.

Pf. of Lem 1. If for some $d\mid n(d\lt n)$, $p\mid x^d-1$, then $p\mid (x^d-1,\Phi_n(x))$. Since $\Phi_n(x)$ does not share any roots with $x^d-1$, we have $\Phi_n(x)\mid\frac{x^n-1}{x^d-1}$. Thus $p\mid(x^d-1,\frac{x^n-1}{x^d-1})$, where $\frac{x^n-1}{x^d-1}=\frac{(x^d-1+1)^{\frac{n}{d}}-1}{x^d-1}=A(n^d-1)+\frac{n}{d}$ (note that this requires $x^d\neq1$). Then $p\mid\frac{n}{d}\mid n$.