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In Beal's Conjecture, are we assuming that we are allowed to repeat numbers? The conjecture is the following:

Beals Conjecture: If $a^n+b^n = c^n $where $n \geq 2$ and $a,b,c,n \in \mathbb{Z}^{+}$ then $a,b$ and $c$ have a common prime factor.

But isn't this basically the opposite of Fermat's Last Theorem? Or does Fermat's Last Theorem require $a,b,c$ to be distinct?

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A generalization of Fermat's last theorem (mathworld.wolfram.com/BealsConjecture.html) –  lab bhattacharjee Jun 6 '13 at 18:30
That's not Beals conjecture you've posted. –  Thomas Andrews Jun 6 '13 at 18:33
Slashdot and Texas Banker, huh? –  Zeta.Investigator Jun 6 '13 at 18:49

2 Answers 2

This is not Beal's conjecture. You have in fact stated Fermat's Last Theorem. Beal's conjecture states that if $a^x + b^y = c^z$, where $a,b,c > 0$ and $x,y,z > 2$, then $\gcd(a,b,c) \neq 1$. Fermat's Last Theorem is an easy corollary of Beal's Conjecture.

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This is not Beal's conjecture! Beal's conjecture is about $$a^x + b^y = c^z$$ where $x,y,z$ are probably not equal.

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