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Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.

(Added Nov 14) The congruence condition $p=21,29$ mod 40 is determined by $-1$ and $5$ are quadratic residue while $2$ is a non residue. Also the sign can be simply predicted by whether $5$ is a 4th power residue mod $p$. For the case $p=21$ mod 40, the sign is $5^{(p-1)/4}$ mod $p$ and the sign is $-(5)^{(p-1)/4}$ mod $p$ if $p=29$ mod 40.

Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.

(Added) The congruence condition $p=21,29$ mod 40 is determined by $-1$ and $5$ are quadratic residue while $2$ is a non residue. The sign depends on whether $5$ is a quartic residue mod $p$, ie $5^{(p-1)/4} = \pm 1$ mod $p$.

Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.

Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.

(Added Nov 14) The congruence condition $p=21,29$ mod 40 is determined by $-1$ and $5$ are quadratic residue while $2$ is a non residue. Also the sign can be simply predicted by whether $5$ is a 4th power residue mod $p$. For the case $p=21$ mod 40, the sign is $5^{(p-1)/4}$ mod $p$ and the sign is $-(5)^{(p-1)/4}$ mod $p$ if $p=29$ mod 40.

Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^9)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.

Not a complete solution.

Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let

$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$

It is known (see for example, Daveport's Multiplicative number theory, p.19),

that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$

where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and

$N_p(x)R_p(x)=(x^p-1)/(x-1)$.

If one compute a few terms of $Z_p(x)$, one can discover

immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$

if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for

these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$),

it will follow that for any primitive 10th root of unity $\eta$,

$R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so

$R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40).

It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum.

We now observe that the very useful trick of writing

$(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely

$R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $

$= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $

Now set $m=10,a=1,p=40k+21$, we get

$R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$

Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with

$r>p/10$ which equals

$(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$

since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue

while 10 is nonresidue. So the number of negative $\sin$ factors is

$(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$.

Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$.

Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal,

we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$.

Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show

$R_p(\zeta^3)=R_p(\zeta)$.

The case for $p=29$ mod 40 is similar.