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

Question

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 "squares subgroup" (?) 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 "squares subgroup" of $\mathbb{F}_p^*$? And why $2 \cdot 3$ must be in this group?

Answer 1

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 $\varphi : \Bbb F_p^* \to \Bbb F_p^*$ given as $x \mapsto x^2$.

In the below, I am also going to assume that $p \notin \{2, 3\}$.

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, $6Q = 3Q \implies 2 = 6 \cdot 3^{-1} \in Q$, 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 $p \neq 2, 3$).

Answer 2

Take the squares $1,4,9,\ldots\pmod p$.
This set forms a subgroup because, if $a^2$ and $b^2$ are in the set, so is their product, because $a^2b^2=(ab)^2$.
Every number in the set appears twice in the list because, if $a^2=b^2\pmod p$ then $a^2-b^2=0\pmod p$, and $p$ is a factor of $(a+b)(a-b)$. So $a^2=(p-a)^2\pmod p$ and there are no other matches. So the set contains $(p-1)/2$ different numbers $\pmod p$.
Multiplication $\pmod p$ shuffles the numbers. We already saw that multiplying by $a^2$ sends one square to another square. So it shuffles the $(p-1)/2$ squares among themselves. So it must shuffle the $(p-1)/2$ non-squares among themselves too.
So the product of a square and a non-square is a nonsquare. Now, when you multiply numbers by a non-square $c$, all the squares become non-squares. That uses up all the nonsquares, so $c$ times the non-squares must become squares. So the product of two non-squares is a square.

Answer 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.