$\rm\:p\equiv 3\pmod 4\:$ is simple for biquadratic residues. This goes back to Gauss, e.g. see Wikipedia.
If prime $\rm\: p\equiv 3 \pmod 4\:$ then every quadratic residue $\rm\:(mod\ p)\:$ is also a biquadratic residue. The first supplement of quadratic reciprocity states $−1$ is a quadratic nonresidue, so for any integer $\rm\:x,\:$ one of $\rm\:x\:$ and $\rm\:−x\:$ is a quadratic residue, and the other a nonresidue. Thus, if $\rm\:r \equiv a^2\pmod p\:$ is a quadratic residue, then if $\rm\:a\equiv b^2\:$ is a residue, $\rm\:r \equiv a^2\equiv b^4 $ is a biquadratic residue, and if $\rm\:a\:$ is a nonresidue, $\rm\:−a\:$ is a residue, $\rm\:−a\equiv b^2,\:$ and again, $\rm\:r \equiv (−a)^2 \equiv b^4\:$ is a biquadratic residue.