# Difference between revisions of "Cohn's criterion"

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If <math>b=2</math>, we will need to prove another lemma: | If <math>b=2</math>, we will need to prove another lemma: | ||

− | All of the zeroes of <math>f(x)</math> satisfy Re <math>z | + | All of the zeroes of <math>f(x)</math> satisfy Re <math>z<\frac{3}{2}</math>. |

Proof: If <math>n=1</math>, then the two polynomials are <math>x</math> and <math>x\pm 1</math>, both of which satisfy our constraint. For <math>n=2</math>, we get the polynomials <math>x^2</math>, <math>x^2\pm x</math>, <math>x^2\pm 1</math>, and <math>x^2\pm x\pm 1</math>, all of which satisfy the constraint. If <math>n\geq 3</math>, | Proof: If <math>n=1</math>, then the two polynomials are <math>x</math> and <math>x\pm 1</math>, both of which satisfy our constraint. For <math>n=2</math>, we get the polynomials <math>x^2</math>, <math>x^2\pm x</math>, <math>x^2\pm 1</math>, and <math>x^2\pm x\pm 1</math>, all of which satisfy the constraint. If <math>n\geq 3</math>, |

## Revision as of 08:03, 4 March 2021

Let be a prime number, and an integer. If 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, is irreducible.

If , we will need to prove another lemma:

All of the zeroes of satisfy Re .

Proof: If , then the two polynomials are and , both of which satisfy our constraint. For , we get the polynomials , , , and , all of which satisfy the constraint. If ,

If Re , we have Re , and then For , then . Therefore, is not a root of .

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.