<s>[INST] Something interesting about the quintic $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0$ and its cousins (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.

(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.
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: 



*

*Is it true that for $B$, there is always an ordering of its roots such that$$\small y_1 y_2 + y_2 y_3 + y_3 y_4 + y_4 y_5 + y_5 y_1 - (y_1 y_3 + y_3 y_5 + y_5 y_2 + y_2 y_4 + y_4 y_1) = 0$$

*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 such quintics and they answer the two questions in the affirmative. But is it true for all prime $p=10m+1$?
 [/INST] I think the depressed quintic in the question is what Emma Lehmer called the reduced quintic in her paper, 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, 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.
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