We have $$x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, 10^k}{11}\Bigr)$$ $$x^5 + x^4 - 12 x^3 - 21 x^2 + x + 5=0,\quad\quad x =\sum_{k=1}^{6}\,\exp\Bigl(\tfrac{2\pi\, i\, 6^k}{31}\Bigr)$$ and so on for prime $p=10m+1$. Let $\alpha$ be this class of quintics with $x =\sum_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, n^k}{p}\Bigr).$ 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 $\alpha$ as $\beta$.

Questions:

1. Is it true that for $\beta$, 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 such quintics and they answer the two questions in the affirmative. But is it true for all prime $p=10m+1$?