Find the number of solutions to $15=a^{2}+b^{2}$.
My professor told us to write $15$ in the form $2^{a}p_{1}^{t_{1}}\cdots p_{n}^{t_{n}}q_{1}^{c_{1}}\cdots q_{m}^{c_{m}}$, and if any $t_{i}$ is odd, then there are no solutions $(a,b)$, and if all $t_{i}$ are even, then there are $4\prod _{i=1}^{m}(c_{i}+1)$ solutions.
My question is: What do $p_{i}$ and $q_{i}$ stand for? I know they are prime numbers, but why are they written using different letters? Is it because all $p_{i} \equiv 3$ (mod $4$) and $q_{i}\equiv 1$ (mod $4$)? But that wouldn't make sense to me because $15=3\cdot 5$ and $3\equiv 3$ (mod $4$) is not true.
Yes, the $p_i$ and $q_i$ are primes and $p_i \equiv 3 \pmod 4$ and $q_i \equiv 1 \pmod 4$. The point is the sum of two squares theorem says that primes $\equiv 3 \pmod 4$ cannot be written as the sum of two squares and implies that numbers which have an odd number of factors of a prime $\equiv 3 \pmod 4$ cannot be written as a sum of two squares. Certainly $3$ cannot. If you do what your teacher says you write for $15\ p_1=1, t_1=1,q_1=5,c_1=1$. Because $t_1=1$ is odd you cannot write $15$ as the sum of two squares and indeed you cannot.