$x^{\frac{p-1}{2}}+1$ is reducible in $\Bbb Z_p[x]$

Question

Let $p$ be an odd prime. Prove that the polynomial $f(x) = x^{\frac{p-1}{2}}+1$ is reducible in $\Bbb Z_p[x]$ and factor $f(x)$ into irreducible polynomials in $\Bbb Z_p[x]$.

I've been struggling this problem. If $\frac{p-1}{2}$ is odd, then $-1$ is zero of $f(x)$, I think there must be a relationship between $x^2+1$ and $x^{\frac{p-1}{2}}+1$ when it comes to $\frac{p-1}{2}$ is even.

My question :

1. Is there a solution when "$x^{\frac{p-1}{2}}$ is even?"
2. How could I show that $f(x)$ is factored into irreducible polynomials?

Answer 1

$f(x) = x^{\frac{p-1}{2}}+1$ factors completely over $\mathbb Z/(p)$: $$ x^{\frac{p-1}{2}}+1 = \frac{x^{p-1}-1}{x^{\frac{p-1}{2}}-1} = \frac{\prod_{a=1}^{p-1} (x-a)}{\prod_{a=1}^{\frac{p-1}{2}} (x-a^2)} $$ This follows from Euler's criterion for quadratic residues.

Answer 2

Over the finite field $\mathbb{F}_p$, the polynomial $x^{p-1}-1$ splits into linear factors, as any nonzero element is a root. Now we have $$ x^{p-1}-1=(x^{\frac{p-1}{2}}-1)(x^{\frac{p-1}{2}}+1), $$ which implies that the two factors also are a product of linear factors over $\Bbb F_p$.