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Im stuck on where to go from what I have. My proof so far is "If p|a-b then a is congruent to b mod p. Then by fermat's last theorem then a^p is congruent to a mod p and same for b^p. This leads to a^p - b^ p being congruent to a - b mod p." but i dunno where to go from there or even if what I have is on the right track. I know what p^2 wont necessarily be a prime but it will still be divisible by p.

Sze
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    Everything you've written is correct but it doesn't tell you much about what is happening $\bmod p^2$. One approach is to try writing $a = kp + b$ and expanding out $a^p - b^p$ and see what you get. Another is to factor $a^p - b^p$. You may also find it helpful to play around with small examples, say $p = 3, 5$. – Qiaochu Yuan Sep 05 '20 at 20:47
  • Does this answer your question? Let $p$ be an odd prime. Prove that if $a\equiv b\pmod{p}$, then $a^p\equiv b^p\pmod{p^2}$. Note if $p=2$, the same procedures in this other post also work. Alternatively, you can use that $a$ and $b$ are both either even or odd, with $a^2$ & $b^2$ then both being $0$ or $1$ modulo $4$, so their difference is a multiple of $p^2 = 2^2 = 4$. – John Omielan Sep 05 '20 at 21:00
  • See also Theorem $\ \ \color{#0a0}{a\equiv b}\pmod{!n},\Rightarrow, a^{\large n}\equiv b^{\large n}\pmod{!n^{\large 2}}\ $ for all integers$\ a,b,n,$ with $,n\ge 0\ \ $ – Bill Dubuque Sep 05 '20 at 21:18

2 Answers2

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Hint: If $p~ | ~a-b$, one may write $$a\equiv b~({\rm mod~}p).\qquad (1)$$ Now $$a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+\cdots+b^{p-1}).$$ Since $p~|~a-b$, it suffices to prove that $$p~|~(a^{p-1}+a^{p-2}b+\cdots+b^{p-1})$$ $$\Leftrightarrow (a^{p-1}+a^{p-2}b+\cdots+b^{p-1})\equiv 0~({\rm mod~}p).$$ Now use (1) to replace $b$ by $a$ and conclude.

Pythagoras
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Since $a^p-b^p=(a-b) (a^{p-1}+a^{p-2}b+...+ab^{p-2}+b^{p-1})$ the end is to show that $a^{p-1}+a^{p-2}b+...+ab^{p-2}+b^{p-1}\equiv 0 \mod p$. Now we have that $a\equiv b \mod p$ and so $a^{p-1}+a^{p-2}b+...+ab^{p-2}+b^{p-1}\equiv b^{p-1}+b^{p-2}b+...+bb^{p-2}+b^{p-1}=pb^{p-1}\equiv 0 \mod p$

Alessandro Cigna
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