Prove that $f(x)=(x^2-2)(x^2-3)(x^2-6)$ has a root in $\mathbb{F}_p$ for every $p$ prime number This is the solution that my book gives ($\mathbb{F}_p$ is the finite field with $p$ elements):

We can assume that $p\neq 2,3$ because $0$ is a root of $f(x)$ in $\mathbb{F}_2$ and $\mathbb{F}_3$. Let $Q$ be the &quot;squares subgroup&quot; (?) of index $2$ in $\mathbb{F}_p^*$. If $2$ and $3$ are not in $Q$ (this means that $(x^2-2)(x^2-3)$ doesn't have roots in $\mathbb{F}_p$), then $2 \cdot 3 \in Q$ (?), and then $x^2-6$ has a root in $\mathbb{F}_p$.

Honestly, I don't understand this solution at all: what is the &quot;squares subgroup&quot; of $\mathbb{F}_p^*$? And why $2 \cdot 3$ must be in this group?
By the squares subgroup, they mean the subgroup $$Q = \{a \in \Bbb F_p^* \mid a = b^2 \text{ for some } b \in \Bbb F_p^*\}.$$
In other words, $Q$ is the image of the homomorphism $\varphi : \Bbb F_p^* \to \Bbb F_p^*.$
Since $\ker \varphi = \{1, -1\}$ and $-1 \neq 1$ (why?), we see that $G$ has index $2$ in $\Bbb F_p^*$.
Now, if $2, 3, 6 \notin Q$, then this means that the cosets $2Q, 3Q, 6Q$ are all not equal to $Q$. Since $Q$ has index $2$ in $\Bbb F_p$, this means that $2Q = 3Q = 6Q$.
But then, $6Q = 3Q \implies 6 \cdot 3^{-1} = 2 \in Q$, a contradiction.
Thus, one of $2, 3, 6$ must be in $Q$. In turn, the given polynomial has a root in $\Bbb F_p$ (in fact, in $\Bbb F_p^*$, if $p \neq 2, 3$).
If 2 is a quadratic residue modulo p, then first factor has a root.
If 3 is a quadratic residue modulo p, then second factor has a root.
If 2 and 3 are quadratic nonresidues modulo p, then 6 is a quadratic residue (because Legendre symbol is multiplicative), therefore third factor has a root.