# Cohn's criterion

Let be a prime number, and an integer. If $\overline{p_np_{n-1}\cdotsp_1}$ (Error compiling LaTeX. ! Undefined control sequence.) is the base- representation of , and , then is irreducible.

## Proof

The following proof is due to M. Ram Murty.

We start off with a lemma. Let . Suppose , , and . Then, any complex root of , , has a non positive real part or satisfies .

Proof: If and Re , note that: This means if , so .

If , this implies if and . Let . Since , one of and is 1. WLOG, assume . Let be the roots of . This means that . Therefore, $f(x) is irreducible.

If$ (Error compiling LaTeX. ! Missing $ inserted.)b=2$, we will need to prove another lemma:

All of the zeroes of$ (Error compiling LaTeX. ! Missing $ inserted.)f(x)z>\frac{3}{2}$.

Proof: If$ (Error compiling LaTeX. ! Missing $ inserted.)n=1xx\pm 1n=2x^2x^2\pm xx^2\pm 1x^2\pm x\pm 1n\geq 3$, <cmath>|\frac{f(z)}{z^n}|\geq |1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^n})>|1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-\frac{1}{|z|^2(|z|-1)}</cmath>

If Re$ (Error compiling LaTeX. ! Missing $ inserted.)z\geq 0\frac{1}{z^2}\geq 0|z|\geq \frac{3}{2}|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1zf(x).

However, if Re , we have from our first lemma, that , so Re . Thus we have proved the lemma.

To finish the proof, let . Since , one of and is 1. WLOG, assume . By our lemma, . Thus, if are the roots of , then . This is a contradiction, so is irreducible.