# Fermat–Catalan conjecture

In number theory, the Fermat–Catalan conjecture combines ideas of Fermat's last theorem and the Catalan conjecture, hence the name. The conjecture states that the equation

${\displaystyle a^{m}+b^{n}=c^{k}\quad }$

(1)

has only finitely many solutions (a,b,c,m,n,k) with distinct triplets of values (am, bn, ck); here a, b, c are positive coprime integers and m, n, k are positive integers satisfying

${\displaystyle {\frac {1}{m}}+{\frac {1}{n}}+{\frac {1}{k}}<1.}$

(2)

This inequality restriction on the exponents has the effect of precluding consideration of the known infinitude of solutions of (1) in which two of the exponents are 2 (such as Pythagorean triples).

As of 2015, the following ten solutions to (1) are known:[1]

${\displaystyle 1^{m}+2^{3}=3^{2}\;}$
${\displaystyle 2^{5}+7^{2}=3^{4}\;}$
${\displaystyle 13^{2}+7^{3}=2^{9}\;}$
${\displaystyle 2^{7}+17^{3}=71^{2}\;}$
${\displaystyle 3^{5}+11^{4}=122^{2}\;}$
${\displaystyle 33^{8}+1549034^{2}=15613^{3}\;}$
${\displaystyle 1414^{3}+2213459^{2}=65^{7}\;}$
${\displaystyle 9262^{3}+15312283^{2}=113^{7}\;}$
${\displaystyle 17^{7}+76271^{3}=21063928^{2}\;}$
${\displaystyle 43^{8}+96222^{3}=30042907^{2}\;}$

The first of these (1m+23=32) is the only solution where one of a, b or c is 1, according to the Catalan conjecture, proven in 2002 by Preda Mihăilescu. While this case leads to infinitely many solutions of (1) (since we can pick any m for m>6), these solutions only give a single triplet of values (am, bn, ck).

It is known by the Darmon–Granville theorem, which uses Faltings' theorem, that for any fixed choice of positive integers m, n and k satisfying (2), only finitely many coprime triples (abc) solving (1) exist;[2][3]:p. 64 but the full Fermat–Catalan conjecture is a much stronger statement since it allows for an infinitude of sets of exponents m, n and k.

The abc conjecture implies the Fermat–Catalan conjecture.[1]

Beal's conjecture is true if and only if all Fermat-Catalan solutions use 2 as an exponent for some variable.