Studying elementary number-theory I found this equivalence for an odd prime $p$:
$(p\equiv1 \,\,\, \text{ (mod 4) and } p\equiv\pm3 \,\,\, \text{ (mod 8)) or } (p\equiv3 \,\,\, \text{ (mod 4) and } p\equiv \pm1 \,\,\, \text{ (mod 8))} \Longleftrightarrow (p\equiv5 \,\,\, \text{ (mod 8) or } p\equiv7 \,\,\, \text{ (mod 8))}$
I there an easy way to see why this equivalence holds? I started prooving $"\Rightarrow":$
If $(p\equiv1 \,\,\, \text{ (mod 4) and } p\equiv\pm3 \,\,\, \text{ (mod 8))}$, then $8\mid2(p-1)$ and $8 \mid (p\pm3)$, so we get $8\mid(2(p-1)-(p\pm3)) \Rightarrow 8\mid p-5 $ and $8\mid p+1 \Rightarrow (p\equiv5 \,\,\, \text{ (mod 8) and } p\equiv7 \,\,\, \text{ (mod 8))}\,\,...$
It seems to take long to show an equivalence like that. Is there an easier way to see why this equivalence holds?
If I understand your question, if you want $p \equiv 1 \pmod 4$ and $p \equiv \pm 3 \pmod 8$ simultaneously then we must have $p \equiv -3 \equiv 5 \pmod 8$, since if instead $p \equiv 3 \pmod 8$ then $p=8k+3$ for some $k$ and we just take this modulo $4$ to see $p \equiv 3 \pmod 4$ which contradicts our original assumption. You can use the same reasoning for the other direction