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.