*(**Update**)*:
Courtesy of Myerson's and Elkies' answers, we find a second cyclic quintic for $\cos\frac{2\pi}{p}$ with $p=\text{1 mod 10}$ as,
$$\frac{z^5}{\beta} = 10 z^3 - 20 n^2 z^2 + 5 (3 n^4 - 25 n^2 - 625) z - 4 n^2 (n^4 - 25 n^2 - 125)$$
where $\beta=n^4 + 25 n^2 + 125.$ Its discriminant is,
$$D=2^{12}\,5^{20}\,(n^4+7n^2)^2\,(n^4 + 25 n^2 + 125)^4$$
Finding a parametric cyclic quintic was one of the aims of this post.




---


*(**Original post**)*:
We have
$$x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0,\quad\quad x =\sum\_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, 10^k}{11}\Bigr)$$
and so on for prime $p=10m+1$. Let $A$ be this class of quintics with $x =\sum\_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, n^k}{p}\Bigr).$ I was trying to find a pattern to the coefficients of this infinite family, perhaps something similar to the Diophantine equation $u^2+27v^2=4N$ for [the cubic case](https://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic).


First, we can *depress* these (get rid of the $x^{n-1}$ term ) by letting $x=\frac{y-1}{5}$ to get the form,
$$y^5+ay^3+by^2+cy+d=0$$
Call the depressed form of $A$ as $B$.


**Questions:**


1. Is it true that for $B$, there is always an ordering of its roots such that$$\small y\_1 y\_2 + y\_2 y\_3 + y\_3 y\_4 + y\_4 y\_5 + y\_5 y\_1 - (y\_1 y\_3 + y\_3 y\_5 + y\_5 y\_2 + y\_2 y\_4 + y\_4 y\_1) = 0$$
2. Do its coefficients $a,b,c,d$ *always* obey the Diophantine relations,
$$a^3 + 10 b^2 - 20 a c= 2z\_1^2$$ $$5 (a^2 - 4 c)^2 + 32 a b^2 = z\_2^2$$ $$(a^3 + 10 b^2 - 20 a c)\,\big(5 (a^2 - 4 c)^2 + 32 a b^2\big) = 2z\_1^2z\_2^2 = 2(a^2 b + 20 b c - 100 a d)^2$$ for integer $z\_i$?


I tested the [first forty](https://math.stackexchange.com/questions/1996552/any-more-cyclic-quintics) such quintics and they answer the two questions in the affirmative. *But is it true for all prime $p=10m+1$?*



Something interesting about the quintic $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0$ and its cousins
I think the depressed quintic in the question is what Emma Lehmer called the reduced quintic in her [paper](https://projecteuclid.org/euclid.dmj/1077476385), The quintic character of 2 and 3, Duke Math. J. Volume 18, Number 1 (1951), 11-18, MR0040338. In the proof of Theorem 4, she writes that the reduced quintic is $$F(z)=z^5-10pz^3-5pxz^2-5p\left({x^2-125w^2\over4}-p\right)z+p^2x-{x^3+625(u^2-v^2)w\over8}$$ where $16p=x^2+50u^2+50v^2+125w^2$, $xw=v^2-u^2-4uv$ determines $x$ uniquely. You might plug these into your conjectured diophantine relations, to see whether they hold. 


**EDIT:**


A slightly different formula is given by [Berndt and Evans](http://www.ams.org/journals/bull/1981-05-02/S0273-0979-1981-14930-2/S0273-0979-1981-14930-2.pdf), The determination of Gauss sums, Bull Amer Math Soc 5, no. 2 (1981) 107-129, on page 119 (formula 5.2). It goes 
$$
F(z)=z^5-10pz^3-5pxz^2-5p\left({x^2-125w^2\over4}-p\right)z+p^2x-p{x^3+625(u^2-v^2)w\over8}
$$
the difference being the factor of $p$ in the last term. Berndt and Evans note that Lehmer inadvertently omitted this factor.


**Update:** (by OP) 


Using the edited coefficients of $F(z)$ by Berndt and Evans, if we express it as,
$$z^5+az^3+bz^2+cz+d=0$$
and taking into account $16p=x^2+50u^2+50v^2+125w^2,\,x=(v^2-u^2-4uv)/w$, then
$$a^3 + 10 b^2 - 20a c =2\times125^2p^2w^2=2z\_1^2$$
$$5(a^2 - 4c)^2 + 32a b^2 = 500^2 p^2(u^2 - u v - v^2)^2=z\_2^2$$
$$(a^3 + 10 b^2 - 20 a c)\,\big(5 (a^2 - 4 c)^2 + 32 a b^2\big) = 2(a^2 b + 20 b c - 100 a d)^2$$
which confirms all three Diophantine relations by the OP.


