Feit-Thompson conjecture: Difference between revisions

From Number
(Created page with '==Statement== For any two distinct primes <math>p,q</math>, the numbers <math>\Phi_p(q) = (q^p - 1)/(q-1)</math> and <math>\Phi_q(p) = (p^q - 1)/(p-1)</math> are relatively prim...')
 
No edit summary
 
Line 1: Line 1:
==Statement==
==Statement==


For any two distinct primes <math>p,q</math>, the numbers <math>\Phi_p(q) = (q^p - 1)/(q-1)</math> and <math>\Phi_q(p) = (p^q - 1)/(p-1)</math> are relatively prime.
For any two distinct primes <math>p,q</math>, the number <math>\Phi_p(q) = (q^p - 1)/(q-1)</math> does not divide <math>\Phi_q(p) = (p^q - 1)/(p-1)</math> are relatively prime.


This conjecture is false, and a counterexample is provided by <math>p = 17, q = 3313</math>.
A stronger form of the conjecture is that the numbers <math>\Phi_p(q)</math> and <math>\Phi_q(p)</math> are relatively prime, and a counterexample to this stronger form is provided by <math>p = 17, q = 3313</math>. This is the only counterexample for <math>p,q < 400000</math>.


==Related facts==
==Related facts and conjectures==


* [[Groupprops:Odd-order implies solvable|Feit-Thompson theorem]]: The Feit-Thompson theorem states that any [[groupprops:odd-order group of odd order]] is [[groupprops:solvable group]]. The proof of this theorem would be considerably simplified if the Feit-Thompson conjecture were true.
* [[Groupprops:Odd-order implies solvable|Feit-Thompson theorem]]: The Feit-Thompson theorem states that any [[groupprops:odd-order group of odd order]] is [[groupprops:solvable group]]. The proof of this theorem would be considerably simplified if the Feit-Thompson conjecture were true.
* [[Goormaghtigh conjecture]]: This conjecture states that the equation:
<math>\frac{x^m - 1}{x - 1} = \frac{y^n - 1}{y - 1}</math>
has only two solution pairs.
==External links==
===Other subject wikis===
* [[Groupprops:Feit-Thompson conjecture]]

Latest revision as of 13:25, 28 April 2009

Statement

For any two distinct primes p,q, the number Φp(q)=(qp1)/(q1) does not divide Φq(p)=(pq1)/(p1) are relatively prime.

A stronger form of the conjecture is that the numbers Φp(q) and Φq(p) are relatively prime, and a counterexample to this stronger form is provided by p=17,q=3313. This is the only counterexample for p,q<400000.

Related facts and conjectures

xm1x1=yn1y1

has only two solution pairs.

External links

Other subject wikis