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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.

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    $\begingroup$ Remember that the Jacobi symbol can equal $1$ even without there being solutions to a congruence, such as $(\frac{-1}{9}) = (\frac{-1}3)^2 = 1$. $\endgroup$ – Greg Martin Aug 12 at 5:15
  • $\begingroup$ Yea, I forgot about that, so is this approach of mine feasible? $\endgroup$ – user799951 Aug 12 at 5:17
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    $\begingroup$ First of all, there's always solutions where $p^a$ divides both $x$ and $y$, for example; perhaps you want to exclude such solutions. The standard way to go from information modulo primes to information modulo their powers is Hensel's lemma. $\endgroup$ – Greg Martin Aug 12 at 5:18
  • $\begingroup$ I have found that p = 1 mod 4 for $x^{2}$ + $y^{2}$ = 0 (mod p) to have solutions but I don't know how to go from that $\endgroup$ – user799951 Aug 12 at 5:22
  • $\begingroup$ If you want solutions where $x$ and $y$ are coprime to $p$, then the admissible $p^a$ are $2$ and $p^a$ for $p\equiv1\pmod4$. $\endgroup$ – Angina Seng Aug 12 at 5:29
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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 $p$.
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