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I am trying to solve the following problem: Let $a$ and $b$ be positive integers and let $p$ be a prime number. Prove that if $a^p \equiv$ $b^p$ (mod $p$), then $a \equiv b$ (mod $p$).

My attempt to solve this starts off by using the idea that $b^p-a^p$ is a multiple of $p$, i.e. $pm=b^p-a^p$, for some $m \in \mathbb Z$. Then I use the idea that what I'm trying to prove is equivalent to $pn=b-a$ for some $n \in \mathbb Z$, because $a \equiv b$ (mod p) means that $b-a$ is a multiple of $p$.

Is this a good start? Also, I was thinking I may have to invoke Fermat's LT at some point, but I am not yet familiar with how Fermat's LT is typically used to solve a problem like this.

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$a^p\equiv a\pmod{p}$ –  André Nicolas Mar 7 at 1:27

2 Answers 2

up vote 2 down vote accepted

Fermat's Little Theorem is exactly what you use! If $p$ is prime, then $a^p \equiv a \pmod{p}$ and $b^p\equiv b \pmod{p}$. So simply, $a^p \equiv a \equiv b^p\equiv b \pmod{p} \implies a\equiv b\pmod{p}$ and you are done.

If you'd like a proof of Fermat's Little Theorem because you say you're not familiar with it yet, I'd be happy to provide one!

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I am sure the internet is teeming with FLT proofs, so I'll do a quick search and see if I can find one that is easy to follow. I appreciate the offer, though, and thanks for the help. –  idkmybffjill Mar 7 at 1:51
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For convenience: primes.utm.edu/notes/proofs/FermatsLittleTheorem.html. And no problem! –  WChang Mar 7 at 1:53
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Thanks. That was very helpful. I should have recognized multiplying by a would yield the corollary, which is what I needed to use. –  idkmybffjill Mar 7 at 2:28

This follows more or less immediately from FLT. Think about what FLT tells you about $a^p$ (mod p) and $b^p$ (mod p). If $a^p\cong b^p$ (mod p) then those two expressions are equivalent by transitivity.

Alternatively, you don't need FLT to prove this at all. Try to factor $a^p-b^p$ and see what you get. $p$ is prime, so what does that tell you about the expression $p|\alpha\beta$?

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