While doing my research on elementary number theory, I came across the following problem which I cannot overcome: Let $p$ be an odd prime, $g$ be any primitive of $p$. Define $$f(p)=\prod_{1\le i <j\le\frac{p-1}{2}}j^2-i^2\pmod p$$ and $$h(p,g)=\prod_{1\le i <j\le\frac{p-1}{2}}g^{2j}-g^{2i} \pmod p .$$ What I want to know is the relationship between $f(p)$ and $h(p,g)$. I calculated the first one hundred primes and find that either $f(p)+h(p,g)=p $ or $f(p)=h(p,g) $. For example $f(17)=4 $ , $h(17,5)=13 $ and $f(73)=h(73,11)=46 $. I believe that this is true for all primes $p$ and all primitives $g$. Now my questions are :

  1. Does it true that we always have $f(p)+h(p,g)=p $ or $f(p)=h(p,g) $
  2. Is it possible to evaluate $f(p)$ and $h(p,g)$ and find the condition when does $f(p)+h(p,g)=p $ and $f(p)=h(p,g) $

    I am eager to know any answer, link, or hints to this problem, thank you!!


I presume you are doing the calculations modulo $p$. In the first case the admissible $i^2$ are the quadratic residues modulo $p$. In the second case the admissible $g_{2i}$ are also the quadratic residues modulo $p$.

Both products have the form $\prod_{1\le i<j\le m}(a_j-a_i)$ where $a_1,\ldots,a_m$ ($m=\frac12(p-1)$) are the distinct quadratic residues modulo $p$. But the ordering of the $a_i$ differs. If the first product is $\prod_{1\le i<j\le m}(a_j-a_i)$ and the second is $\prod_{1\le i<j\le m}(b_j-b_i)$, then $b_j=a_{\tau(j)}$ for some permutation $\tau\in S_m$ and then $\prod_{1\le i<j\le m}(b_j-b_i)=\text{sgn}(\tau) \prod_{1\le i<j\le m}(a_j-a_i)$ (at least modulo $p$). Here $\text{sgn}(\tau)$ is the sign of the permutation $\tau$.

  • $\begingroup$ yes , you are right. I omitted "mod p". Now I have corrected that $\endgroup$ – 王李远 Jul 27 '18 at 15:26
  • $\begingroup$ thank you very very much! Can you determine the signature of $\tau$? I am waiting for you answer!!! $\endgroup$ – 王李远 Jul 27 '18 at 15:32
  • $\begingroup$ @王李远 You can try to generalize the result from this question: mathoverflow.net/questions/302865 $\endgroup$ – punctured dusk Jul 27 '18 at 15:44
  • $\begingroup$ @barto Thanks for you advice.I will try to generalize that method $\endgroup$ – 王李远 Jul 27 '18 at 15:55

We are working in the field $F=\Bbb F_p$ for a prime $p$. Let $g$ be a multiplicative generator, a generator of the cyclic group $(\Bbb F_p^\times, \cdot)$. Then $g$ is also generating the $-1$ in the field, and of course $g^{p-1}=1$, $g^{(p-1)/2}=-1$.

Now let us consider the set $S$ of the elements $1,2,\dots,(p-1)/2$ considered in $F$. These are representatives of $F^\times/\{\pm 1\}$. The class of $k$ is covered by $k, p-k$. Taking the squares, we obtain a list of all squares in $F$.

And also the set $T$ of the elements $g^1,g^2,\dots,g^{(p-1)/2}$ considered in $F$. These are also representatives of $F^\times/\{\pm 1\}$. The class of $k=g^u$ (for a suitable unique $u$) is covered by $k=g^u,\ -k=g^{u+(p-1)/2}$. (And $u$, $u+(p-1)/2$ are not both in the list.)

Taking squares, we have as sets the equality $\{ j^2\ :\ 1\le j\le (p-1)/2\ \}=\{\ g^u\ :\ 1 \le u\le (p-1)/2\ \}$, but for the products in the OP we also have to use the orderings of the sets, as inherited by the ones of the parametrizing indices $j$, respectively $u$. The products differ then by the sign of the reordering.

  • $\begingroup$ ... oh, same as above, i'm definitively too slow in typing. $\endgroup$ – dan_fulea Jul 27 '18 at 15:40
  • $\begingroup$ any way , thanks for your answer, which is also useful to me!! $\endgroup$ – 王李远 Jul 27 '18 at 15:57

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