TITLE: Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
QUESTION [7 upvotes]: 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.)

REPLY [7 votes]: 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.