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(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:

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

(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.

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

(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:

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

(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:

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

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 $x =\sum_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, n^k}{p}\Bigr).$ 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 $\alpha$ 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 $p=10m+1$?

(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:

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