 Solution

The modular congruence $x^4\equiv 1$ first implies $x\not\equiv 0$. Multiplication by such an $x$ permutes the nonzero residue classes modulo the prime $p$: if $xa\equiv xb$, cancellation gives $a\equiv b$. Multiply all these residues and cancel their nonzero product to obtain $x^{p-1}\equiv 1$, the usual short proof of Fermat little theorem.

Now $(x^2-1)(x^2+1)\equiv 0$. By the Euclid lemma, either $x^2\equiv 1$ or $x^2\equiv -1$. In the latter case, because $p-1=4k+2$,contradicting the preceding product argument, since $p$ is odd. ThusNo unproved assertion about the number of fourth roots is needed.