7
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Zhi-Wei Sun Nov 1 '18 at 12:21
  • 1
    $\begingroup$ Foe what I said in the previous comments, see Lemma 4.1 of my preprint arxiv.org/abs/1810.12102. $\endgroup$ – Zhi-Wei Sun Nov 2 '18 at 1:43
6
$\begingroup$

Here is a respectively short way to write down what we came up with Dmitry Krachun tonight.

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 $2^m\equiv (-1)^{(p^2-1)/8}=(-1)^{(m+1)/2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.