Let $n$ have the prime power factorization $n=2^e p_1^{a_1}\cdots p_k^{a_k}q_1^{b_1}\cdots q_l^{b_l},$ where the $p_i$ are distinct primes of the form $4t+1$, and the $q_i$ are distinct primes of the form $4t+3$.
If one or more of the $b_i$ is odd, then $n$ has no representations as a sum of two squares. If all the $b_i$ are even, then the number $N(n)$ of representations is given by $N(n)=4(a_1+1)(a_2+1)\cdots (a_k+1).$ However, this counts all representations, including the ones that use negative integers. It also considers $0^2+5^2$ as different from $5^2+0^2$.
If we don't want negatives, and order doesn't matter, things are more complicated. If $8$ divides $N(n)$, then the number of essentially distinct representations is equal to $\frac{N(n)}{8}.$
If $N(n)$ is not divisible by $8$, it is of the form $4q+4$. Then the number of representations is $\frac{N(n)+4}{8}.$
Remark: Let $D_1$ be the number of positive divisors of $n$ of the form $4t+1$, and $D_3$ the number of positive divisors of $n$ of the form $4t+3$. It is a nice result of Jacobi that $N(n)=4(D_1-D_3).$ Unfortunately, I do not see how this can be used to give a more efficient algorithm for computing $N(n)$.