Is there any way to express in $g$ the number $b$ for which $b^2 \equiv -1 mod g$ when g is a product of primes $\equiv 1 mod 4$?
So, first of all, can $b$ be expressed in terms of $p \equiv 1 mod 4$, where $b^2 \equiv -1 mod p$?
Obviously, both $b$ and $-b$ will do, but my question is whether $b$ can be expressed in $p$ in general.
Then, for $n = \prod_{j=1}^{k} p_j^{n_j}$, can $b$ be expressed in terms of $p_j \equiv 1 mod 4$, where $b^2 \equiv -1 mod n$?
You need to find a solution $\bmod$ every prime power factor of $g$ and then stitch them together using the chinese remainder theorem/algorithm.
However, if there is a prime $p$ that is congruent to $3\bmod 4$ then it is impossible, this is because if $g$ is a primite root $\bmod p$ then $-1=g^{(p-1)/2}$, and this number cannot be a square if $(p-1)/2$ is odd.
It is also impossible for $4$ to divide $g$.
So $g$ is of the form $p_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$ or $2p_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$ where every prime is congruent to $1\bmod 4$.
Now, there is no known easy way to find a solution to $x^2\equiv -1 \bmod p^a$ ( it is easy to show that there are exactly two solutions, because a primitive root exists). The best way is probably to look for a primitive root. testing whether a number $r$ is a primitive root can be done in time $\log(p^a)^2$ by checking if $r^{p-1/q}$ is $1\bmod p^k$ for each prime $q$ dividing $\varphi(p^k)$. The probability you pick a generator each time is $\frac{p-1}{p}$, so this method is pretty fast. After you have a generator your answer is $g^{(p-1)/4}$ and $g^{3(p-1)/4}$.