On the partner of the Emma Lehmer quintic

Question

Given, $$x^5+10cx^3+10dx^2+5ex+f = 0$$ If there is an ordering of its roots such that, $$\small x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0\tag1$$ then its coefficients are related by the quadratic in $f$, $$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2 $$ This implies that such quintics come in pairs, having the same $c,d,e$ but differing only in $f$. An example would be the solvable Emma Lehmer quintic, $$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} +1=0$$ and its partner (also solvable) $$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} - (n^3 + 5n^2 + 10n + 20)(n^3 + 5n^2 + 10n + 5)/25=0$$

Q: Is there a subset of $n$ such that the partner also has Galois order 5?

Answer 1

[Edited to give more details]

The subset of such $n \in \bf Q$ is empty: as long as the "partner" quintic, call it $Q$, is irreducible, its Galois group is cyclic of order $5$ over ${\bf Q}(\sqrt{5})$, but dihedral of order $10$ over $\bf Q$.

Let the roots of $Q$ be $x_1,x_2,x_3,x_4,x_5$, ordered so that there is a Galois automorphism $\sigma$ taking each $x_i$ to $x_{i+1}$ (with $x_6 \equiv x_1$). Then if $Q$ is irreducible over some field $K$ then its Galois group is cyclic if and only if $$ \Delta := (x_1 - x_2) (x_2 - x_3) (x_3 - x_4) (x_4 - x_5) (x_5 - x_1), $$ $$ \Delta' := (x_1 - x_3) (x_3 - x_5) (x_5 - x_2) (x_2 - x_4) (x_4 - x_1) $$ are in $K$ (because $\Delta,\Delta'$ are invariant under $\sigma$, but taken to $-\Delta,-\Delta'$ by the involution $x_i \leftrightarrow x_{6-i}$). Now $\Delta \Delta' = \pm \cal D$, where ${\cal D}^2$ is the discriminant of $Q$; we compute ${\cal D} = D_3 D'_3 D_4^2 / 5^3$, where $$ D_3 = 3 n^3 + 20 n^2 + 50 n + 50, \quad D'_3 = 4 n^3 + 10 n^2 + 25 n + 25, $$ $$ D_4 = n^4 + 5 n^3 + 15 n^2 + 25 n + 25. $$ Moreover, $\Delta$ and $\Delta'$ are integral over ${\bf Q}[n]$, and even over ${\bf Z}[\frac15][n]$ with bounded denominator; so there are not many possibilities for the factorization ${\cal D} = \Delta \Delta'$, and we can determine the correct one by specializing $n$ and computing the $x_i$ numerically. We find that $\Delta$ and $\Delta'$ are $\pm 5^{-3/2} D_3 D_4$ and $\pm 5^{-3/2} D_3' D_4$ (or vice versa if we change $x_1,x_2,x_3,x_4,x_5$ to $x_1,x_3,x_5,x_2,x_4$). These are always rational over ${\bf Q}(\sqrt 5)$, but never over $\bf Q$ (for rational $n$). Therefore the Galois group is always cyclic over ${\bf Q}(\sqrt 5)$ but dihedral over $\bf Q$. $\Box$