Validity of this number theoretic expression? Background
I was working on a long winded idea and I think the below should be correct:
If there are $2$ primes $p_i$ and $p_j$ such that $p_i \neq 2 \neq p_j$. Then we define $\omega_{p_i}= e^{i \pi/p_i}$
Now do $p_i$ and $p_j$ obey the below?
$$ \prod_{m=1}^{p_i -1 } \frac{(1+\omega_{p_i}^m)}{(1-\omega_{p_i}^m)^2} \prod_{n=1}^{p_j -1} \frac{(1+\omega_{p_j}^n)}{(1-\omega_{p_j}^n)^2} = \prod_{l \neq m}(A_l - A_m)$$
Where $A_l$ can be anything within the set $(\omega_{p_i}, \omega^2_{p_i}, \omega^3_{p_i}, \dots \omega^{p_i -1}_{p_i}, \omega_{p_j}, \omega^2_{p_j}, \omega^3_{p_j}, \dots \omega^{p_j -1}_{p_j})$
Question
Is the above (conjecture) correct (can you prove it) ? Or provide a simple counter example?
 A: (Too long for a comment.)
Let $p$ be an odd prime and $\omega=e^{i \pi / p}$, then $\omega^p=-1$ and $\omega^2$ is a primitive $p^{th}$ root of unity, so:

*

*$P(z)=\prod_{k=1}^{p-1}\left(z-\omega^{2k}\right)=\dfrac{1-z^p}{1-z}=1+z+z^2+\dots+z^{p-1}$


*$\prod_{k=1}^{p-1}\left(1-\omega^{2k}\right)=P(1) = p$


*$\prod_{k=1}^{p-1}\left(1-\omega^{-2k}\right)=\overline{\prod_{k=1}^{p-1}\left(1-\omega^{2k}\right)}=\overline{P(1)} = p$
Let $U=\prod_{k=1}^{p-1}\left(1+\omega^k\right)$ then:
$$
\require{cancel}
\begin{align}
U^2 &= \prod_{k=1}^{p-1}\left(1+\omega^k\right)\left(1+\omega^{p-k}\right)
\\ &= \prod_{k=1}^{p-1}\left(\bcancel{1}+\bcancel{\omega^p}+\omega^k+\omega^p\omega^{-k}\right)
\\ &= \prod_{k=1}^{p-1}\omega^{k}\left(1 - \omega^{-2k}\right)
\\ &= \omega^{p(p-1)/2}\,P(1)
\\ &= \omega^{p(p-1)/2} \, p
\end{align}
$$
Let $V=\prod_{k=1}^{p-1}\left(1-\omega^k\right)$ then:
$$
\begin{align}
V^2 &= \prod_{k=1}^{p-1}\left(1-\omega^k\right)\left(1-\omega^{p-k}\right)
\\ &= \prod_{k=1}^{p-1}\left(\bcancel{1}+\bcancel{\omega^p}-\omega^k-\omega^p\omega^{-k}\right)
\\ &= \prod_{k=1}^{p-1}\omega^{-k}\left(1 - \omega^{2k}\right)
\\ &= \omega^{-p(p-1)/2}\,P(1)
\\ &= \omega^{-p(p-1)/2} \, p
\end{align}
$$
Then the LHS of the proposed identity is $\;\displaystyle\frac{U_1}{V_1^2}\,\frac{U_2}{V_2^2}\,$. This leaves open the question of choosing the square root branches for $U_1,U_2$ but may still be a step towards an explicit evaluation.
A: I wrote a python script to test this, and it failed for the first case I tried, $p_i=3,p_j=5$
Here's the script:
from sympy import E, re, im, I, pi, latex, expand
from itertools import combinations
from functools import reduce

def product(seq):
    return reduce(lambda x,y:x*y, seq, 1)

def test(p,q):
    # Pre: p and q are distinct odd primes
    wp = E**(I*pi/p)
    wq = E**(I*pi/q)
    ppowers = [wp**m for m in range(1,p)]
    qpowers = [wq**m for m in range(1,q)]
    np = product([1+ w for w in ppowers])
    nq = product([1+ w for w in qpowers])
    dp = product([(1- w)**2 for w in ppowers])
    dq = product([(1- w)**2 for w in qpowers])
    A = combinations(ppowers+qpowers, 2)
    Ap = -product([(a-b)**2 for a,b in A])
    s = expand(Ap*dp*dq-np*nq)
    print(latex(re(s)),'+\\left(', latex(im(s)),'\\right) i')
    print(float(re(s)), float(im(s)))

test(3,5)

The test function clears denominators in the given expression, and returns the difference of the right- and left hand sides, so that if the conjecture is true, the return value should be $0$.
I got the result
$$
- 8400 \sqrt{3} \sqrt{\frac{\sqrt{5}}{8} + \frac{5}{8}} - 12000 \sqrt{3} \sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}} + 3300 \sqrt{5} + 18675 +\left( - 2 \sqrt{3} \left(\frac{1}{4} - \frac{\sqrt{5}}{4}\right) + 2 \sqrt{3} \left(\frac{1}{4} + \frac{\sqrt{5}}{4}\right) \right) i
$$
or approximately $$0.00033524787724058605+ 3.872983346207417i$$
Unless I've misinterpreted something, or made some careless programming error, the conjecture is false.
