Catalan's conjecture: Difference between revisions

Latest revision as of 17:14, 13 August 2010

This conjecture states that the solution set to Catalan's Diophantine problem:

$x^{p}-y^{q}=1$ ,

for $x,y$ positive integers not equal to $0,1$ and $p,q$ positive integers greater than one, has precisely one solution: $x=3,y=2,p=2,q=3$ .

The conjecture has been proved.

Conjecture	Statement of conjecture	Status
Fermat-Catalan conjecture	$a^{m}+b^{n}=c^{k}$ has only finitely many solutions for $a,b,c$ positive integers and ${\frac {1}{m}}+{\frac {1}{n}}+{\frac {1}{k}}<1$	open
abc conjecture	For every $\epsilon$ , there exists $C_{\epsilon }$ such that if $a+b=c$ , then $\max\{\|a\|,\|b\|,\|c\|\}\leq C_{\epsilon }\prod _{p\|abc}p^{1+\epsilon }$ , the product over $p$ prime

@@ Line 5: / Line 5: @@
 <math>x^p - y^q = 1</math>,
-for <math>x,y \ne 0,1</math> and <math>p,q</math> positive integers greater than one, has ''precisely'' one solution: <math>x = 3, y = 2, p = 2, q = 3</math>.
+for <math>x,y</math> positive integers not equal to <math>0,1</math> and <math>p,q</math> positive integers greater than one, has ''precisely'' one solution: <math>x = 3, y = 2, p = 2, q = 3</math>.
 The conjecture has been proved.
+==Relation with other facts/conjectures==
+{| class="sortable" border="1"
+! Conjecture !! Statement of conjecture !! Status
+|-
+| [[Fermat-Catalan conjecture]] || <math>a^m + b^n = c^k</math> has only finitely many solutions for <math>a,b,c</math> positive integers and <math>\frac{1}{m} + \frac{1}{n} + \frac{1}{k} < 1</math> || open
+|-
+| [[abc conjecture]] || For every <math>\epsilon</math>, there exists <math>C_{\epsilon}</math> such that if <math>a + b = c</math>, then <math>\max \{ |a|, |b|, |c| \} \le C_\epsilon \prod_{p | abc} p^{1 + \epsilon}</math>, the product over <math>p</math> prime ||
+|}