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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, as well as this one, and this one.

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For question 1, the answer is yes, as shown by Emma Lehmer herself. (See the paper here, 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 would imply that $P(n)$ does take on prime values infinitely often.

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  • $\begingroup$ Wonderful! There's actually a formula for $a,b$. Very interesting paper, by the way. $\endgroup$ – Tito Piezas III Dec 17 '14 at 17:42

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