Quintic reciprocity conjecture

Let $p=x^4 + 25x^2 + 125$ be a prime. Prove that $2$ is a quintic residue $\pmod p$, and therefore $y^5=2\pmod p$ is solvable.

A similar example was first conjectured by Euler:

If $p=x^2 + 27$ is a prime, then $2$ is a cubic residue $\pmod p$, $y^3=2\pmod p$ is solvable.

UPDATE:

Digging into this more, my conjecture holds for $x < 1000$ plus more values, and has many common properties with Euler's cubic reciprocity conjecture involving the polynomial $x^2 + 27$.

According to this page on cubic reciprocity, $2$ is a cubic residue $\pmod p$ if and only if $p=x^2 + 27y^2$.

Similarly, if $p=x^4 + 25x^2y^2 + 125y^4$, then $2$ is a quintic residue $\pmod p$. This also seems to hold.

Besides $2$ being a cubic and quintic residue of $x^2 + 27$ and $x^4 + 25x^2 + 125$, respectively, $x$ also holds the same property.

If $p=x^2 + 27$, then $x$ is a cubic residue $\pmod p$.

If $p=x^4 + 25x^2 + 125$, then $x$ is a quintic residue $\pmod p$.

Euler's logic and reasoning about his conjectures on cubic residues could likely be applied to quintic residues.

Solution 1:

This is a special case of a result due to Emma Lehmer (see this article). In (8), set $u = 0$, $v = 4t$ and $w = 4$ (observe that $x$, $u$ and $v$ are even, so $2$ is a quintic residue modulo $p$); then $16p = 16t^4 + 50 \cdot 16t^2 + 125 \cdot 16$, and division by $16$ produces the polynomial $p = t^4 + 50t^2 + 125$. I'm sure that a suitable choice of the parameters will produce your polynomial.