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:*** 


1. 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 can ***always*** be given in the form of $(1)$ with integer $a,b,c$?
2. Also, does $P(n)$ assume prime values infinitely often?


**P.S.** This was inspired by cubic analogues I asked about [in this MSE post](https://math.stackexchange.com/questions/1065862/), as well as [this one](https://math.stackexchange.com/questions/1068084/), and [this one](https://math.stackexchange.com/questions/1069520/).



Does the Lehmer quintic parameterize certain minimal polynomials of the $p$th root of unity for infinitely many $p$?
For question 1, the answer is yes, as shown by Emma Lehmer herself. (See [the paper here](http://www.ams.org/journals/mcom/1988-50-182/S0025-5718-1988-0929551-0/home.html), in particular, equation (5.8) on page 539.) In particular, Lehmer states that one can take 
$$
 a = \frac{\left(\frac{n}{5}\right) - n^{2}}{5}, \quad b = \left(\frac{n}{5}\right).
$$
(Here $\left(\frac{n}{5}\right)$ denotes the Legendre symbol.) This polynomial is defining the unique degree $5$ subfield of $\mathbb{Q}(\zeta\_{p})$ and so we take $c$ to be any element in $\mathbb{F}\_{p}^{\times}$ of order $\frac{p-1}{5}$.


Question 2 is definitely open. (In fact, it is not known if there is a polynomial $P(n)$ of degree $> 1$ that takes on prime values infinitely often.) [Bunyakovsky's conjecture](http://en.wikipedia.org/wiki/Bunyakovsky_conjecture) would imply that $P(n)$ does take on prime values infinitely often.


