Is it acceptable to say that if $a^b ≡ a^c\;(mod\;p)$, then $b≡c\;(mod\;p)\;?$ I am trying to write a proof, and I want to know if this is an acceptable assumption to make.
 A: No, it's not. For example,
$$1^2\equiv 1^3\mod 5$$
but obviously $2\not\equiv3\mod 5$.

If you'd tried this out with any number $a\not\equiv0\mod p$, you'd have found that $a^b\equiv a^c\mod p$ pretty much never means $b\equiv c\mod p$.

I cannot stress this enough, but try numerical examples to your theorem before you try to prove it. It gives you a good idea of how problems work, how you could prove it, and in this case, if the statement is even true.
A: No it is not. However, you can do the following. Define $\text{ord}_p(a)$ to be smallest positive integer $d$ satisfying
$$
a^d \equiv 1\pmod {p}.
$$
It is known that for any other positive integer $n$ for which $a^n \equiv 1 \pmod {p}$, we must have $d \mid n$. To see why, note that if $n = kd + q$, for $0 < q \leq d-1$, we have
$$
a^n \equiv a^{kd+q} \equiv a^q \equiv 1 \pmod {p} \implies a^q\equiv 1 \pmod {p}
$$
contradicting with the fact that $d$ is the smallest of all such exponents.
Now assuming $p$ is a prime and $(p,a) = 1$, you can rewrite your congruence as 
$$
a^{b-c} \equiv 1 \pmod {p}.
$$
Hence, you can say $b\equiv c\pmod{\text{ord}_p(a)}$.
A: As follows from Little Fermat Theorem, for any $a$, $$a^p \equiv_p a$$
Therfore, if $$a^b \equiv_p a^b$$
, then
$$a^b \cdot a^p \equiv_p a^c \cdot a$$
$$a^{b+p} \equiv_p a^{c+1}$$
If your assertion was true, then both $b \equiv_p c$ and $b \equiv_p b + p \equiv_p c +1$, which is a clear contradiction. 
