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After reading an article by Peter Norvig about Beal's Conjecture I became pretty interested in digging into it. In his article here he goes into the use of a technique to optimize checking by this:

If $A^x + B^y = C^z$ and $p$ is some word-sized prime then $A^x \pmod p + B^y \pmod p = C^z \pmod p$.

I am aware of a few different properties of modular arithmetic with primes that are useful, but I do not see how this one is derived. I have a feeling that it involves the Chinese Remainder Theorem and some fiddling, but I couldn't find a proof of this equality anywhere and it's really starting to bother me.

Could someone tell me how Norvig arrived at this conclusion? I would really appreciate it. I've never seen such an equality before. Thank you!

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There’s nothing relevant about the exponents used in this equation or pertinent about the modulus being prime. To clarify matters, let $a=A^x,$ $b=B^y,$ and $c=C^z.$ If $a+b=c$, then obviously $(a+b) \equiv c\bmod p$ (in the notation Norvig is using, this is saying $(a+b)(\bmod p)=c(\bmod p)).$ So, the only question you might have is whether $ a(\bmod p) + b(\bmod p) = (a+b)(\bmod p).$ But this is clear since if $r_a \equiv a\bmod p$ and $r_b\equiv b \bmod p,$ then we know that $(r_a + r_b)\equiv (a+b)\bmod p.$

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