A guess about prime number and modular arithmetic.

I have following guess:

Let $p>2$ be a prime number. Then there exist an integer $1 < a < p$ such that for any two different integer $x, y \in [p-1]$, we have $a^x \ne a^y \pmod{p}$.

I tried the first few examples, like $p=5, a = 3$, and $p=7, a=5$, and $p=19, a=2$, all fulfills the above conditions. So is there anything number theory could say something about this?

Answer

There are deep reasons that this is true, but the fundamental reason is that the integers modulo $p$ form something called a field, and, in fields, there can be no more than $n$ solutions to the equation $x^{n}-1=0$ for any $n$. This can be used to show that there must be a solution to $x^{p-1}-1=0$ which is not a solution to any $x^{d}-1$ for $d

This $a$ is called a "generator" because when you list $a,a^2,a^3,\dots,a^{p-1} \pmod p$, you get $p-1$ distinct values which thus cover all of the nonzero elements, modulo $p$.