Question on modular arithmetic? Would this ever be possible:
$$a b^2 ≡ a \pmod p$$
where $p$ is a prime number and $1&lta&ltp$ and $1&ltb$
Please back your answer with some kind of proof other than Fermat's Little Theorem. This isn't homework.
 A: If $a\equiv0\bmod p$, the given relation is true for all $b$.
If $a\not\equiv0\bmod p$ then $a$ can be cancelled from both sides of the congruence which reduces to $b^2\equiv1\bmod p$, hence $b\equiv\pm1\bmod p$.
A: Hint $\ $ prime $\rm\: p\ |\ ab^2-a = a\:(b-1)(b+1)\ \Rightarrow\ p\ |\ a\ \ or\ \ p\ |\ b-1\ \ or\ \ p\ |\ b+1$
A: Since $p$ is prime, we can cancel $a$ from both sides unless $a\equiv 0 \pmod p$ (this doesn't always work if $p$ is a composite number).  We get $b^2 \equiv 1 \pmod p$.  Again, since $p$ is prime, a number can have only two mod-$p$ square roots (it can have more than two in some cases if $p$ is composite).  The two square roots of $1$ are $\pm 1$ and $-1$ is of course the same as $p-1$.
So $ab^2\equiv a\pmod p$ holds if either $a\equiv 0$ or $b\equiv\pm1$, but not otherwise.
So which parts of this do you want proofs of?  The proof that nonzero numbers are invertible in mod $p$ (thereby justifying the cancelation from both sides)?  I once posted an answer that shows how to find the inverse.  The proof that there can't be more than two square roots if $p$ is prime?  (That latter fact is a sort of corollary.)
