There is a lot of fascination with the [Vandermonde determinant](https://en.wikipedia.org/wiki/Vandermonde_matrix) and for many good reasons and purposes. My current quest is more of number-theoretic.



> 
> **QUESTION.** Let $p\equiv 3$ (mod $4$) be a prime. What is the value of
>  $$\prod\_{1\leq i<j\leq\frac{p-1}2}(j^2-i^2) \mod p?$$
> 
> 
> 



Vandermonde determinant: modulo
I saw this in [this recent arxiv paper](https://arxiv.org/abs/1809.07766) by Zhi-Wei Sun (Quadratic residues and related permutations and identities, version 6, equation (1.5)). I'll reproduce the proof in case it goes under further revisions. Every equality below is $\pmod{p}$. Start by factoring


$$\prod\_{1\le i<j\le \frac{p-1}{2}}(j^2-i^2)=\prod\_{1\le i<j\le \frac{p-1}{2}}(j-i)(j+i)$$
and noticing that
$$\prod\_{1\le i<j\le \frac{p-1}{2}}(j-i)=\prod\_{k=1}^{\frac{p-1}{2}}k^{\frac{p-1}{2}-k}$$
$$\prod\_{1\le i<j\le \frac{p-1}{2}}(j+i)=\prod\_{k=1}^{\frac{p-1}{2}}k^{\lfloor\frac{k-1}{2}\rfloor}(p-k)^{\lfloor\frac{k}{2}\rfloor}=(-1)^{\sum\_{k=1}^{\frac{p-1}{2}}\lfloor k/2\rfloor}\prod\_{k=1}^{\frac{p-1}{2}}k^{k-1}=\prod\_{k=1}^{\frac{p-1}{2}}k^{k-1}$$
Therefore
$$\prod\_{1\le i<j\le \frac{p-1}{2}}(j^2-i^2)=\left(\frac{p-1}{2}\right)!^{\frac{p-1}{2}-1}=1$$


