Conjecture on cubic-root values of the quadratic-residue product Sp(x)S_p(x)

From papers

Let Sp(x)S_p(x) be the polynomial defined in the source, let ω\omega be a primitive third root of unity, let p>3p>3 be a prime with p3(mod4)p\equiv3\pmod4, and let h(p)h(-p) be the class number of the imaginary quadratic field of discriminant p-p. Write (p)\left(\frac{\cdot}{p}\right) for the Legendre symbol, and let (xp,yp)(x_p,y_p) be the least positive integer solution of

3x2+4(p3)=py2.3x^2+4\left(\frac p3\right)=py^2.

Then

The cubic-root product conjecture.

Sp(ω±1)=(1)(h(p)+1)/2(p3)xp3ypp2×{i±1if p7(mod12),(1){1k<p3:(kp)=1}(iω)±1if p11(mod12),S_p(\omega^{\pm1})=(-1)^{(h(-p)+1)/2}\left(\frac p3\right)\frac{x_p\sqrt3\mp y_p\sqrt p}{2}\times\begin{cases}i^{\pm1}&\text{if }p\equiv7\pmod{12},\\(-1)^{|\{1\le k<\frac p3:(\frac kp)=1\}|}(i\omega)^{\pm1}&\text{if }p\equiv11\pmod{12},\end{cases}

and

Sp(ωˉ)=(1)(h(p)1)/2(p3)xp3+ypp2×{iif p7(mod12),(1){1k<p3:(kp)=1}iωˉif p11(mod12).S_p(\bar\omega)=(-1)^{(h(-p)-1)/2}\left(\frac p3\right)\frac{x_p\sqrt3+y_p\sqrt p}{2}\times\begin{cases}i&\text{if }p\equiv7\pmod{12},\\(-1)^{|\{1\le k<\frac p3:(\frac kp)=1\}|}i\bar\omega&\text{if }p\equiv11\pmod{12}.\end{cases}

The source presents this as an open conjecture about exact values of Sp(x)S_p(x) at roots of unity; examples are given for p=79p=79 and p=227p=227.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

Zhi-Wei Sun, “Trigonometric identities and quadratic residues”, arXiv:1908.02155 (2021).

Solutions 0

No solutions have been posted yet.