Showing that $(x^2 - 2)(x^2 - 3)(x^2 - 6)$ has a root in $\mathbb{F}_p$ Let $p$ be a prime number, $K = \mathbb{F}_p$ the field with $p$ elements, and $f = (x^2 - 2)(x^2 - 3)(x^2 - 6) \in K[x]$. I now want to show that $f$ has a root in $K$.
I know that to show the statement, it suffices to show that there exists an $a \in K$ so that at least one of these factors in $f$ becomes $= 0$ for that $a$. But I don't know how I can generally show that. I tried out (and verified) the statement for the first few prime numbers, hoping that this would give me a general idea, but I haven't been successful so far.
I was given the hint that I could at some point use the fact that $K^\times$ is cyclic, but I don't see how that can come into play.
 A: If $a\in K^\times$ is not a quadratic residue, then the bijection $\mu_a:K^\times\to K^\times$ given by multiplication by $a$ takes any quadratic residue to a non-residue, and therefore any non-residue to a residue (since exactly half of $K^\times$ are quadratic residues; this may be proven using cyclicity). Now note that $\mu_2(3)=6$.
Reformulation using cyclicity more explicitly: Let $a$ be a generator of the group $K^\times$. Then $2=a^m,3=a^n$ and $6=2\cdot3=a^{m+n}$. At least one of the exponents $m,n$ and $m+n$ must be even, which means one of $2,3$ and $6$ must be a square in $K$.
A: By the multiplicativity of Legendre symbol 
$$ \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{ab}{p}\right)\tag{1} $$
it follows that for every prime $p\geq 5$ at least one element of the set $\{2,3,6\}$ is a quadratic residue: if both $2$ and $3$ are quadratic non-residues, $6$ is a quadratic residue. It follows that at least one polynomial among $x^2-2,x^2-3,x^2-6$ completely splits over $\mathbb{F}_p$. The cases $p=2$ and $p=3$ are trivial.
