When a square of a polynomial root is in $\mathbb{F}_p$ Let $p > 5$ be prime. If $1 + 5x^2 + 5x^4$ has a root in $w \in \mathbb{F}_{p^2}$, then does $-1 + x + x^2$ have a root in $\mathbb{F}_p$? If $w$ is a root of the former, then $5w^2 + 2$ is a root of the latter, so it is sufficient to show that $w^2 \in \mathbb{F}_p$. I believe we can establish the claim if $w^p = \pm w$ because then $(w^2)^p = (w^p)^2 = (\pm w)^2 = w^2$. Since in this case $w^2$ is fixed by $x \mapsto x^p$, it is in $\mathbb{F}_p$. Next note that $x \mapsto x^p$ permutes the roots of the polynomial and $-w$ is another root. I'm not sure how to show that $w$ doesn't get sent to one of the other two roots though. I'm also not sure if there's a simpler way to approach this question.
 A: Let $p > 5$ be prime.

Claim:$\;$If $5x^4+5x^2+1$ has a root in $\mathbb{F}_{p^2}$, then $x^2+x-1$ has a root in $\mathbb{F}_p$.

Proof:

Let $f=5x^4+5x^2+1$, and let $v$ be a root of $f$ in $\mathbb{F}_{p^2}$.

Then $-v$ is also a root of $f$, so we can write $f=5(x^2-v^2)g$ where $g\in\mathbb{F}_{p^2}[x]$ is monic of degree $2$.

It follows that $f$ factors in $\mathbb{F}_{p^2}[x]$ as
$$
f=5(x^2-v^2)(x^2-w^2)
$$
where $w^2\in\mathbb{F}_{p^2}$.

Letting $u\in\mathbb{F}_{p^2}$ be a root of $x^2-5$, we have $u^2=5$, hence
$$
f=u^2(x^2-v^2)(x^2-w^2)
$$
Then from
$$
u^2(x^2-v^2)(x^2-w^2)
=
5x^4+5x^2+1
$$
we get $(uvw)^2=1$, so $uvw=\pm 1$, and then since $u,v\in\mathbb{F}_{p^2}$, it follows that $w\in\mathbb{F}_{p^2}$.

Hence  $f$ splits completely in $\mathbb{F}_{p^2}[x]$ as
$
f=5(x-v)(x+v)(x-w)(x+w)
$.

As you noted, if $v^2\in\mathbb{F}_p$ or $w^2\in\mathbb{F}_p$, the conclusion would follow, so assume $v^2,w^2\not\in\mathbb{F}_p$.

Note that $f$ can't be irreducible over $\mathbb{F}_p$ since $f$ has degree $4$ but all roots of $f$ are in $\mathbb{F}_{p^2}$.

Since $v^2,w^2\not\in\mathbb{F}_p$, it follows that $f$ has no roots in $\mathbb{F}_p$, so $f$ must have a monic factor $h$ of degree $2$ in $\mathbb{F}_p[x]$.

But $x^2-v^2,x^2-w^2\not\in\mathbb{F}_p[x]$.

It follows that the product of the roots of $h$ is $vw$ or $-vw$, hence $vw\in\mathbb{F}_p$.

Then from
$$
5^2(x^2-v^2)(x^2-w^2)
=
5(5x^4+5x^2+1)
$$
we get $(5vw)^2=5$, so $5$ is a square in $\mathbb{F}_p$.

Thus the discriminant of $x^2+x-1$ is a square in $\mathbb{F}_p$, hence by the quadratic formula, it follows that $x^2+x-1$ has a root in $\mathbb{F}_p$.
