*(**Update**)*:
Courtesy of Myerson's and Elkies' answers, we find a second *simple* cyclic quintic for $\cos\frac{\pi}{p}$ with $p=10m+1$ as,
$$F(z)=z^5 - 10 p z^3 + 20 n^2 p z^2 - 5 p (3 n^4 - 25 n^2 - 625) z + 4 n^2 p(n^4 - 25 n^2 - 125)=0$$
where $p=n^4 + 25 n^2 + 125$. Its discriminant is
$$D=2^{12}5^{20}(n^2+7)^2n^4(n^4 + 25 n^2 + 125)^4$$
Finding a simple parametric cyclic quintic was one of the aims of this post.

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*(**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][1].

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][2] such quintics and they answer the two questions in the affirmative. ***But is it true for all prime $p=10m+1$?***


  [1]: http://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic
  [2]: http://math.stackexchange.com/questions/1996552/any-more-cyclic-quintics