# Difference between revisions of "Cohn's criterion"

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This means <math>g(z)>0</math> if <math>|z|\geq \frac{1+\sqrt{1+4H}}{2}</math>, so <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>. | This means <math>g(z)>0</math> if <math>|z|\geq \frac{1+\sqrt{1+4H}}{2}</math>, so <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>. | ||

− | If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible. | + | If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, \phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible. |

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

## Revision as of 08:29, 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 .

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.