# Is this stipulation on Beals conjecture used?

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 ?

• Speed up searching for what? Commented Jun 6, 2021 at 15:28
• 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. Commented Jun 6, 2021 at 15:39
• @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 Commented Feb 18 at 0:08