Assume for a second the Beal's conjecture is correct. It's then true that $A=pa$,$B=pb$, $C=pc$ and the exponent on each gives back $$p^xa^x+p^yb^y=p^zc^z$$ but this shows that if one of $x,y,z$ is minimal, we get back that two terms have $p$ still after division, but the third won't defying distributivity of multiplication over addition/subtraction. The only way to eliminate $p$ completely is if exponents on $p$ in the factorization contain the other exponents ( think parenthesized exponentiation rules) .

Is this ever used to speed up searching ?

  • $\begingroup$ Speed up searching for what? $\endgroup$
    – hardmath
    Commented Jun 6, 2021 at 15:28
  • 1
    $\begingroup$ The stipulation is that $A,B,C$ don’t share a prime factor? Certainly the conjecture is false without it, as $2^3 + 2^3 = 2^4$, for example. $\endgroup$
    – mjqxxxx
    Commented Jun 6, 2021 at 15:39
  • $\begingroup$ @mjqxxxx Beals says all solutions do have a shared factor. I'm saying for any prime except 2 you can't eliminate it from just 1 of the terms. So somehow with coprime x,y,z you need p raised to the same exponent $\endgroup$ Commented Feb 18 at 0:08


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