I have tried using FLT but I didn't really go anywhere from there. The first obvious observation that $x \equiv y\pmod{p}$, and that $x^{p(p+1)} \equiv y^{p(p+1)}\pmod{p^2}$. A hint would be great because I am definitely missing something obvious.
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2$x=kp+y$ then using binomial formula. – Ivan Kaznacheyeu Jul 31 '22 at 07:03
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2Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read its title. – José Carlos Santos Jul 31 '22 at 07:06
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@IvanKaznacheyeu how would i know that it is specifically divisible by $p^2$? could you please elaborate a bit? using the result from $p|x-y$ and binomial theorem, as you said, i've gotten an equation for $x^p - y^p$ but i don't know how to proceed (apologies for the earlier ping) – Jul 31 '22 at 07:15
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Does this answer your question? Prove that if $a^p-b^p$ is divisible by $p$, then it is also divisible by $p^2$ - found through using an Approach0 search, with there being other very similar or duplicate questions, e.g., Show $a^p \equiv b^p \mod p^2$. – John Omielan Jul 31 '22 at 07:15
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1$(kp+y)^p-y^p={p\choose p} k^p p^p+\ldots+{p\choose 2}k^2p^2 y^{p-2}+{p\choose 1}kp y^{p-1}+y^p-y^p$. All summands before $y^p$ are divisible by $p^2$. – Ivan Kaznacheyeu Jul 31 '22 at 07:20
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ah, i missed the $p$ factor in the $kp$ that makes it divisible. thanks for the help Ivan. – Jul 31 '22 at 07:22