Simple Fermat's Little Theorem Corollary

Question

I'm probably missing something super simple, but I just can't figure this out. For integers $a,p$ such that $gcd(a,p)=1$ then Fermat's Little Theorem states $$a^{p-1}=1 \mod{p}$$ My textbook then says if $a$ is a primitive root and $x\equiv y \mod{p-1}$ then $a^x=a^y\mod p$. Can someone help me understand this proof?

Answer 1

You have

$$x \equiv y \pmod{p - 1} \implies x = y + n(p-1) \tag{1}\label{eq1A}$$

for some $n \in \mathbb{N}$. Thus, using $a^{p-1} \equiv 1 \pmod{p}$, you have that

$$a^x \equiv a^{y + n(p-1)} \equiv (a^y)(a^{n(p-1)}) \equiv a^y(a^{p-1})^n \equiv a^y \pmod{p} \tag{2}\label{eq2A}$$

Note this is true for all $a$ where $\gcd(a,p) = 1$, not only for primitive roots of $p$.