If $a^b\equiv c^b\mod p$, can we conclude that $a\equiv c\mod p$? If $a^b\equiv c^b\mod p$, is is true that $a\equiv c\mod p$, where $b$ is odd and $p$ is prime?
We know that if $a\equiv c\mod p$, then $a^b\equiv c^b\mod p$. Is the reverse true?
 A: As the other answers pointed out, the reverse is generally false.
However, if $b$ and $p-1$ are coprime, then $a^b \equiv c^b \bmod p$ implies $a \equiv c \bmod p$.
Proof. If $a \equiv 0 \bmod p$ or $c \equiv 0 \bmod p$ the result is clear (because $p$ is prime), so we will assume that $a \not\equiv 0 \bmod p$ and $c \not\equiv 0 \bmod p$. In this case, Fermat's little theorem shows that $a^{p-1} \equiv 1 \bmod p$ and $c^{p-1} \equiv 1 \bmod p$.
Therefore, for all integers $u,v$, the relation $a^b \equiv c^b \bmod p$ implies
$$
(a^b)^u(a^{p-1})^v \equiv (c^b)^u(c^{p-1})^v \bmod p
$$
By Bezout's theorem we can find integers $u,v$ such that $bu + (p-1)v = 1$, so what we have actually proven is $a^1 \equiv c^1 \bmod p$.

The condition that $b$ and $p-1$ are coprime is actually necessary and sufficient if $p$ is a prime number.
Indeed, if $b$ and $p-1$ are not coprime, we can find integers $b',k$ and $d \geq 2$ such that $b=b'd$ and $p-1 = kd$. Then for all $x$ which is a generator of the cyclic group $(\Bbb Z/p\Bbb Z)^\times$, the integers $a = x^k$ and $c = 1$ will satisfy
$$
a^b \equiv x^{kb'd} \equiv (x^{p-1})^{b'} \equiv 1 \equiv c^b \bmod p
$$
but $a \not\equiv c \bmod p$ because $x^k \equiv 1 \bmod p$ would imply $(p-1) \mid k$, which contradicts $0 < k < p-1$.
A: Hint: Try out some examples with $p=7$ and $b=3$. (The important aspect is that $b\mid p-1$, because $p-1$ is the size of the group $(\mathbb{Z}/p\mathbb{Z})^\times$.)
A: $$2^3\equiv  1^3\mod7 $$but  $$2 \not\equiv  1\mod7$$
A: $2^3=1^3 \mod 7$ and  $2 \neq 1\mod 7$ but $(2)=(1)$ as ideals since we are in a field. So if you mean $(a), (b)$ as ideals the assertion is true since we have only one nonzero ideal in a field. But if you mean $(a)=(b)$ as elements there are counter examples.
