For which prime powers are there solutions to $x^{2} + y^{2} = 0 \pmod {p^a}$ with $(x,p) = (y,p) = 1$?

Question

I have a problem about Jacobi and Legendre Symbols and don't know how to approach it properly since I am new to this subject. For this problem I tried to show that $\left(\dfrac{-1}{p^a}\right) = 1$ and got lost so any help would be awesome.

Answer 1

I'm assuming $xy\neq 0$. As I see it, there are three cases: $p\equiv 1 \pmod 4$, $p=2$, $p\equiv 3 \pmod 4$.

1. $p\equiv 1 \pmod 4$: Using Gaussian integers, one can show $X^2+Y^2=p$ has a solution. Then the Diophantus identity $$ (a^2+b^2)(c^2+d^2)=(ac+bd)^2(ad-bc)^2 $$shows the set of integers expressible as a sum of two squares is closed under multiplication, whence the result follows.
2. $p=2$: Clearly $(x,y)=(1,1)$ is a solution for $a=1$. For $a>1$ there are clearly no solutions because you've specified $(x,2)=(y,2)=1$, which can't be satisfied modulo $4$.
3. $p\equiv 3 \pmod 4$: If $a$ is odd, there are no solutions because $x^2+y^2 \equiv 0,1,2\pmod 4$ but $p^a\equiv 3\pmod 4$. If $a$ is even, there are no solutions either because as @Greg Martin mentioned, the Jacobi symbol yields a valid solution only if every prime factor gives $1$, but we showed there is no solution mod $p$.