The solvable ** Emma Lehmer quintic** is given by,

$$F(y) = y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y + 1 = 0$$

with discriminant $D = (7 + 10 n + 5 n^2 + n^3)^2(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$.

For prime $p=25 + 25 n + 15 n^2 + 5 n^3 + n^4$, we solve $F(y)=0$ in radicals as a sum of powers of the root of the unity $\zeta_p = e^{2\pi i/p}$,

$$y = a+b\sum_{k=1}^{(p-1)/5}\,{\zeta_p}^{c^k}\tag1$$

for integer $a,b,c$. The *complete* table for small $n$,

$$\begin{array}{|c|c|c|c|c|} n &p &a &b &c \\ -1& 11& 0& +1& 10\\ +1& 71& 0& +1& 23\\ -2& 11& -1& -1& 10\\ +2& 191& -1& -1& 11\\ -3& 31& -2& -1& 6\\ -4& 101& -3& +1& 32\\ +4& 941& -3& +1& 12\\ -6& 631& -7& +1& 24\\ +7& 5051& -10& -1& 7\\ -9& 3931& -16& +1& 11\\ \end{array}$$

*Questions:*

- Is it true that for all
*prime*$P(n)=25 + 25 n + 15 n^2 + 5 n^3 + n^4$, then a root of $F(y)=0$ in radicals canbe given in the form of $(1)$ with integer $a,b,c$?*always* - Also, does $P(n)$ assume prime values infinitely often?

**P.S.** This was inspired by cubic analogues I asked about in this MSE post, as well as this one, and this one.