I. Comparison
It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,
$$x^3-nx^2+(n-3)x+1 = 0$$
has the nice property that its roots $a,b,c$, if in correct order, obey,
$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$
(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)
Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,
$$x^5 + n^2x^4 – (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$
It turns out its roots $x_k$ may have a similar property.
II. Question:
Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?
Update 1: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,
$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$
Or more generally (for $\mu$ an integer),
$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde} = 0$$
where $\mu^5$ is the constant term of the quintic and the Lehmer quintic the special case $\mu=1$.
Update 2: I've already tested the Hashimoto septic which fortunately has a seventh power $\mu^7$ as its constant term and the analogous relation,
$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde}-\frac{\mu^{6}}{abcdef}+\frac{\mu^{7}}{abcdefg} = 0$$
and among the $(p-1)! = 720$ permutation of its roots, at least 9 works.
III. Example
Let $n=-1$. Then we get the quintic for $p=11$ and its roots,
$$x^5 + x^4 – 4x^3 – 3x^2 + 3x + 1 = 0$$
$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$
with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.
P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?
Best Answer
The map $s(r) = \frac{n+2 + nr - r^2}{1 + (n+2)r}$ cyclically permutes the roots. This map is given in [2], and I found it through the reference in [1]. It turns out to give the correct order.
Explicitly, if we pick one root $x_1$ we have
\begin{eqnarray*} D &=& n^3 + 5n^2 + 10n + 7 \\ Dx_2 &=& (n^2 + 4n + 4)x_1^4 + (n^4 + 4n^3 + 4n^2 - n - 2)x_1^3 + \\&& (-2n^5 - 14n^4 - 43n^3 - 76n^2 - 80n - 39)x_1^2 + \\&& (n^6 + 9n^5 + 37n^4 + 89n^3 + 131n^2 + 107n + 36)x_1 + \\&& (n^4 + 8n^3 + 26n^2 + 39n + 22) \\ % Dx_3 &=& (-2n - 3)x_1^4 + (-2n^3 - 4n^2 - 3n - 2)x_1^3 + \\&& (3n^4 + 14n^3 + 31n^2 + 41n + 24)x_1^2 + \\&& (-n^5 - 7n^4 - 21n^3 - 36n^2 - 29n - 6)x_1 + \\&& (-2n^2 - 7n - 6) \\ % Dx_4 &=& (-n^2 - 3n - 2)x_1^4 + (-n^4 - 3n^3 - 2n^2 + n + 1)x_1^3 + \\&& (2n^5 + 12n^4 + 33n^3 + 54n^2 + 53n + 23)x_1^2 + \\&& (-n^6 - 8n^5 - 29n^4 - 62n^3 - 81n^2 - 59n - 18)x_1 + \\&& (-n^5 - 7n^4 - 24n^3 - 47n^2 - 52n - 25) \\ % Dx_5 &=& (n + 1)x_1^4 + (n^3 + 2n^2 + 3n + 3)x_1^3 + \\&& (-n^4 - 4n^3 - 9n^2 - 14n - 8)x_1^2 + \\&& (-n^4 - 7n^3 - 19n^2 - 29n - 19)x_1 + \\&& (n^4 + 6n^3 + 16n^2 + 20n + 9) \end{eqnarray*}
and it is pure computation to show that $$x_1 x_2 x_3 x_4 x_5 - x_2 x_3 x_4 x_5 + x_3 x_4 x_5 - x_4 x_5 + x_5 = 0$$
Online computation with sagecell.sagemath.org
However, this isn't a very satisfying proof because it doesn't address the question of necessary and sufficient properties of the original polynomial. Does a suitable permutation of roots exist for all irreducible polynomials of odd (or maybe odd prime) degree whose Galois group is cyclic? The equivalence between the forms
$$ (x_1^4 x_2^3 x_3^2 x_4)^{1/5} + (x_2^4 x_3^3 x_4^2 x_5)^{1/5} + (x_3^4 x_4^3 x_5^2 x_1)^{1/5} + (x_4^4 x_5^3 x_1^2 x_2)^{1/5} + (x_5^4 x_1^3 x_2^2 x_3)^{1/5} = 0 \\ \frac1{x_1} - \frac1{x_1 x_2}+\frac1{x_1 x_2 x_3} - \frac1{x_1 x_2 x_3 x_4} + \frac1{x_1 x_2 x_3 x_4 x_5} = 0 \\ x_1 x_2 x_3 x_4 x_5 - x_2 x_3 x_4 x_5 + x_3 x_4 x_5 - x_4 x_5 + x_5 = 0 $$ and a couple of other variants relies on the constant coefficient of Emma Lehmer's quintic being $1$. If there is a generalisation, which variant does it require? Or does generalisation require a cyclic Galois group and a unit constant coefficient?
[1] Henri Darmon, Note on a polynomial of Emma Lehmer, Math. Comp. vol. 56, no. 194, April 1991, pp. 795-800.
[2] Rene Schoof and Lawrence C. Washington, Quintic polynomials and real cyclotomic fields with large class numbers, Math. Comp. vol. 50, no. 182, April 1988, pp. 543-556.