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If $$a^p\equiv b^p \pmod p$$ where $p$ is prime prove that $$a^p\equiv b^p \pmod{p^2}$$ that problem was at my exam today on number theory and i just didnt have a clear mind to solve it.Although i had alot of thoughts couldnt get them together.

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  • $\begingroup$ This has been asked and answered many times before on MSE. $\endgroup$ – Bill Dubuque Jun 23 '14 at 18:42
  • $\begingroup$ Sorry but i am not aware of what MSE stands for $\endgroup$ – Manolis Lyviakis Jun 23 '14 at 18:44
  • $\begingroup$ MSE is this site Math.StackExchange $\endgroup$ – Bill Dubuque Jun 23 '14 at 18:55
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The first equivalence implies that $a\equiv b\pmod{p}$. There are more clever ways of getting to the latter, but here's a direct method: Writing $b = a + np$ for some integer $n$, we have $$b^p \equiv (a + np)^p \equiv a^p + a^{p-1} n p \binom{p}{1} + \sum_{k \geq 2} \binom{p}{k} a^{n-k} (np)^k \equiv a^p + a^{p-1} n p^2 \equiv a^p \pmod{p^2}$$

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Hint: By Fermat's Theorem, $a^p\equiv a\pmod{p}$ and $b^p\equiv b\pmod{p}$. So $a=b+kp$ for some integer $k$. It follows that $$a^p=(b+kp)^p.$$ Expand the right-hand side using the Binomial Theorem.

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