Let $a\in \mathbb Z$ be such that $a=b^2+c^2,$ where $b,c\in \mathbb Z-\{0\}.$ Then $a$ can be written as
$(1)pd^2,$ where $d\in \mathbb Z$ and $p$ is a prime with $p\equiv1\pmod 4$
$(2)pd^2,$ where $d\in \mathbb Z$ and $p$ is a prime with $p\equiv3\pmod 4$
$(3)pqd^2,$ where $d\in \mathbb Z$ and $p,q$ are primes with $p\equiv1,\pmod 4$,$q\equiv3\pmod 4$
$(4)pqd^2,$ where $d\in \mathbb Z$ and $p,q$ are distinct primes with $p,q\equiv3\pmod 4$
Counter-Example for option 2
$5=2^2+1^2$ but $5\not\equiv 3(mod 4)$
Counter-Example for option 3
$25=3^2+4^2,25=5.5.1^2,5\equiv1(mod4)$ but $5\not\equiv3(mod4)$
What is the explicit counterexample for option (4)?
Also,how to prove option (1)?
Due to the Sum of two squares theorem for (2), (3) and (4) you anyway would have the left side ($a$) to have a factor $p^{2k+1}$ in its prime decomposition. So (2), (3) and (4) cannot represent $a$.