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For a prime $p\equiv 1\pmod 4$, let $\left(\frac{\cdot}{p}\right)_4$ denote the rational biquadratic residue symbol; that is, $$ \left(\frac{a}{p}\right)_4 = \begin{cases} \ \ \ 1\ &\text{if $a$ is a biquadratic residue modulo $p$}, \\ -1\ &\text{otherwise.} \end{cases} $$ Also, denote by $\left(\frac{\cdot}{p}\right)$ the quadratic residue (Legendre) symbol.

Burde has shown that if $p$ and $q$ are primes satisfying $\left(\frac pq\right)=1$ and both congruent to $1$ modulo $4$, then writing $p=a^2+b^2$ and $q=c^2+d^2$ with $a$ and $c$ odd (and $b$ and $d$ even), one has $$ \left(\frac{p}{q}\right)_4\left(\frac{q}{p}\right)_4=\left(\frac{ac-bd}{q}\right). $$

Is there a version of Burde's result for the case where $q\equiv 3\pmod 4$? To be more specific,

Suppose that $p\equiv 1\pmod 4$ is prime, and write $p=a^2+b^2$ with $a$ odd (and $b$ even). Is there any natural way to associate with every prime $q\equiv 3\pmod 4$ satisfying $\left(\frac{p}{q}\right)=1$ integers $c$ and $d$ so that $$ \left(\frac{q}{p}\right)_4=\left(\frac{ac-bd}{q}\right)? $$ (Inserting in the right-hand side factors like $(-1)^{(q-1)/4}$ would be just fine with me.)

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  • $\begingroup$ An answer is given in the wikipedia page you linked in the section under Dirichlet. If $p\equiv \sigma^2 \pmod q$ then $(\frac{-q}{p})_4=(\frac{\sigma(b+\sigma)}{q})$. A reference is given to Lemmermeyer's book. $\endgroup$ – Lucia Nov 23 '14 at 16:05
  • $\begingroup$ @Lucia: I noticed this, but there seems to be a problem with this assertion as if, say, $p\equiv 1\pmod q$, then one can choose both $\sigma=1$ and $\sigma=-1$, while I cannot see any reason for $\left(\frac{b+1}q\right)$ and $\left(\frac{-(b-1)}q\right)$ to be equal to each other. I do not have Lemmermeyer's book handy to check the assertion. $\endgroup$ – Seva Nov 23 '14 at 16:29
  • $\begingroup$ The sign doesn't matter. If you multiply $(\frac{\sigma(b+\sigma)}{q})$ and $(\frac{-\sigma(b-\sigma)}{q})$ together, you get $(\frac{-\sigma^2(b-\sigma^2)}{q}) = (\frac{a^2}{q})=1$. $\endgroup$ – Lucia Nov 23 '14 at 16:31
  • $\begingroup$ @Lucia: How about $q=3$, $p=13=3^2+2^2$, $\sigma=1$, in which case $b=2$ and $q\mid b+1$? Also, is it implicitly assumed that $p\equiv 1\pmod 8$ (this condition is mentioned a couple of lines above the equality in question)? $\endgroup$ – Seva Nov 23 '14 at 17:11
  • $\begingroup$ Seva: It doesn't seem to me that $p$ is assumed to be $1\pmod 8$. When $q$ divides $a$ (which is the example you raised) then the biquadratic symbol is determined by $(\frac{2}{q})$. I'm not an exper, and it may be safest to consult Lemmermeyer's book. $\endgroup$ – Lucia Nov 23 '14 at 17:31
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It is perhaps not a direct analogue, but the reciprocity formula of K. S. Williams, K. Hardy and C. Friesen published in 1985 gives an expression for all odd primes $q>1$. This formula comprises all known rational quartic reciprocity laws:

Theorem (Williams, Hardy, Friesen): Let $p\equiv 1(4)$ be a prime, and let $A,B,C$ be integers such that $A^2=p(B^2+C^2)$, $2\mid B$, $(A,B)=(A,C)=(B,C)=1$, $A+B\equiv 1(4)$. Then for every odd prime $q>1$ with $(p/q)=1$ we have $$ \left(\frac{q}{p}\right)_4=\left(\frac{A+B\sqrt{p}}{q}\right). $$ An elementary proof and a discussion how to derive Burde's law from this is given in Franz Lemmermeyer's article Rational quartic reciprocity.

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