# On $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)$ and $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)$ modulo a prime $p>3$

QUESTION: Is my following conjecture true?

Conjecture. Let $$p>3$$ be a prime and let $$h(-p)$$ be the class number of the imaginary quadratic field $$\mathbb Q(\sqrt{-p})$$. Then

$$\frac{p-1}2!!\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)\equiv \begin{cases}(-1)^{(p-1)/4}\pmod p&\text{if}\ p\equiv 1\pmod4, \\(-1)^{(h(-p)+1)/2}\pmod p&\text{if}\ p\equiv 3\pmod 4.\end{cases}$$ Also, $$\frac{p-3}2!!\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)\equiv \begin{cases}1\pmod p&\text{if}\ p\equiv1\pmod4,\\(-1)^{(p-1+2h(-p))/4}\pmod p&\text{if}\ p\equiv3\pmod4.\end{cases}$$

I have checked the conjecture via a computer. It should be true in my opinion. Your comments are welcome!

• I have proved that the two congruences are equivalent and that the square of each left-hand side is congruent to $1$ modulo $p$. So it remains to determine the signs. Nov 1 '18 at 12:21
• Foe what I said in the previous comments, see Lemma 4.1 of my preprint arxiv.org/abs/1810.12102. Nov 2 '18 at 1:43

Denote $$p=2m+1$$.
The idea is very simple: calculate the product $$\prod_{j\in\{s,s+1\}, 1\leqslant i\leqslant m,\atop p\nmid 2i+j} (2i+j).$$ Note that this is a product of all non-zero residues modulo $$p$$ except $$s+1$$, thus it equals $$-1/(s+1)$$. Now apply this observation for $$s=1,3,\dots,m-1$$ (if $$m$$ is even) and $$s=2,4,\dots,m-1$$ (if $$m$$ is odd) and multiply, you almost get your double product.
Namely, if $$m$$ is even (so $$p\equiv 1\pmod 4$$) you get the whole double product, which appears to be congruent $$\pmod p$$ to $$(-1)^{m/2}/m!!$$ as conjectured.
If $$m$$ is odd, you get that the double product is congruent $$\pmod p$$ to $$\prod_{i=1}^{m-1} (1+2i)\cdot (-1)^{(m-1)/2}/m!!$$ (the first product corresponds to the case $$j=1$$), and since the right hand side of your formula is just $$m!$$ by Mordell, we need only to prove that $$\prod_{i=1}^{m-1} (1+2i)\cdot (-1)^{(m-1)/2}\equiv m!$$, that is easy: write each $$1+2i$$ as $$-2(m-i)$$, you get $$2^{m-1}(m-1)! (-1)^{(m-1)/2}$$ in the left hand side and $$m!=-(m-1)!/2$$ in the right hand side, so we need $$2^m (-1)^{(m+1)/2}\equiv 1$$ which is clear as $$2^m\equiv (-1)^{(p^2-1)/8}=(-1)^{(m+1)/2}$$.