# Expressing quartic Dirichlet characters modulo primes $p\equiv 1\bmod 4$ with Legendre symbols

Looking through some old notes of mine from two years ago I found some crude notes writing what amounted to the statement that for any prime $p\equiv 1\bmod 4$ one could express for any odd integer $p\nmid n$ both of the order four dirichlet characters $\chi_1,\chi_2:(\mathbb{Z}/p\mathbb{Z})^{*}\to \{\pm 1,\pm i\}$ at $n$ as follows:

$$\chi_1(n)=i^{\frac{1-\left(\frac{n}{p}\right)}{2}}(-1)^{\frac{(n-1)(p-1)}{8}}\prod_{k=0}^{\frac{n-1}{2}}\left(\frac{\left\lfloor pk/n\right\rfloor!}{p}\right)$$ $$\chi_2(n)=(-i)^{\frac{1-\left(\frac{n}{p}\right)}{2}}(-1)^{\frac{(n-1)(p-1)}{8}}\prod_{k=0}^{\frac{n-1}{2}}\left(\frac{\left\lfloor pk/n\right\rfloor!}{p}\right)$$

However I can't locate the rest of what I was writing on the scrap paper and I'm not sure what I did. Though here are some things I tried, with a statement I was able to show is equivalent at the end:

First letting $\zeta=-\left(\frac{p-1}{2}\right)!$ gave me that $\zeta^2\equiv -1\bmod p$ with $\left(\frac{\zeta}{p}\right)=\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{4}}$ and so I could write $(x-\zeta y)(x+\zeta y)\equiv x^2+y^2\bmod p$ as well as $\left(\frac{x^2+y^2}{p}\right)=\left(\frac{x-\zeta y}{p}\right)\left(\frac{x+\zeta y}{p}\right)$ however this stopped looking useful when I realized that $\mathbb{Z}[i]\to (\mathbb{Z}/p\mathbb{Z})^{*}[i]$ loses the property of being an integral domain when $p\equiv 1\bmod 4$ as the Gaussian integers form a UFD and we have$(a^2+b^2)(x^2+y^2)=(ax-by)^2+(ay+bx)^2$ However the floor functions appearing in the products of both the identities reminded me of the similar looking floor function sums that appear in Eisenstein's lattice point proof of quadratic reciprocity so following up with that idea I tried a similar approach of partitioning out the quartic residues modulo $p$ based on parity yet this wasn't useful so I tried manipulating the character identities, working backwards with latice sums finding if I wrote:

$$s_p(a)=p^2\sum_{k=1}^{\frac{p-1}{2}}\left\{\frac{k^2}{p}\right\}\left\{\frac{ak^2}{p}\right\}$$

Then the original identities on quartic dirichlet characters become equivalent to the proposition:

$$(-a)^{\frac{p-1}{4}}\equiv \zeta^{\frac{1-\left(\frac{a}{p}\right)}{2}}(-1)^{s_p(a)}\bmod p$$

However I'm not sure how to prove this either, so any help in the manner would be appreciated.

• For $n=1$, $\chi_1(1)=1$, while the RHS of your first equation is $\left( \frac{((p-1)/2)!}{p} \right)$. For $p=5$, for instance, this Legendre symbol is $-1$ ($2$ is not a square modulo $5$), which contradicts your identity. – Ofir Gorodetsky Feb 13 '18 at 4:24
• @OfirGorodetsky I tried translating what I wrote from my notes under a substitution to make it more general, it seems I made an error manipulating the floor function. I updated it in its original form for the restricted case on odd integers, with the product varying over a smaller interval. Though the identity still may be false as I'm quite messy and might of made an error, I think it should hold in this unaltered form. As some other related stuff in the same pile I've verified holds, also I checked it for the first $200$ primes $p\equiv 1\bmod 4$ with odd integers $n$ between $1$ and $p-2$. – Ethan Feb 13 '18 at 4:59
• Do you want any proof or do you want to complete your idea of the proof? This, of course, if your identity is true. If you want any proof, have you tried to show that your formulas are multiplicative? – EFinat-S Feb 14 '18 at 13:32
• @EFinat-S I found the rest of my notes and I wrote out a proof but its like five pages long and doesn't make use of anything particularly "deep" just tedious combinatoral arguments and algebraic manipulation, the first page and a half involves some arguments about atleast one primitive root $g$ modulo $p$ existing such that powers of $g^4$ lie within certain intervals then the next two pages involve manipulating a bunch of floor function and fractional part sums then I use two different identities about finite rectangular lattices. – Ethan Feb 14 '18 at 14:22
• For example here is one of them, (the other identity is too long to put here and its incredibly ugly) given any $m,n,a,b\in \mathbb{N}$ with $\gcd(a,b)=1$ and an arbitrary map $f:[0,n]\times [0,m]\cap \mathbb{Z}\times \mathbb{Z}\to \mathbb{C}$ we have that: $$\sum_{k=0}^{\left\lfloor\min\left(n/a,m/b\right)\right\rfloor}f(ka,kb)+\sum_{v=0}^{m}\sum_{u=0}^{n}f(u,v)=\sum_{v=0}^{m}\sum_{u=0}^{\lfloor \text{min}(av/b,n)\rfloor}f(u,v)+\sum_{u=0}^{n}\sum_{v=0}^{\lfloor \text{min}(bu/a ,m)\rfloor}f(u,v)$$ – Ethan Feb 14 '18 at 14:22