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.)