Let $p$ be a prime of the form $p = a^2 + b^2$ with $a,b \in \mathbb{Z}$ and $a$ an odd prime. Prove that $(a/p) =1$

Question

Let $p$ be a prime of the form $p = a^2 + b^2$ with $a,b \in \mathbb{Z}$ and $a$ an odd prime. Prove that $(a/p) =1$

Could anyone give me a hint for the solution please?

Answer 1

Hint: Note that $p\equiv 1\pmod 4$, so quadratic reciprocity gives $\left(\frac{a}p\right)=\left(\frac{p}a\right)$.

Answer 2

To explain further, we have that $p\equiv b^2 \bmod a$ from the given equation.

Quadratic reciprocity tells us that if $a$ and $p$ are odd primes and either leaves remainder $1$ on division by $4$ we have $p$ is a square $\bmod a$ if and only if $a$ is a square $\bmod p$. (and if both leave remainder $3$ modulo $4$ then precisely one of the primes is a square modulo the other).

Legendre Symbols are a convenient way of writing this - a notation - but it is important to understand what they mean. The fact that $a$ is an odd prime tells us that $p\gt 2$ is odd, and the fact that $p$ is the sum of two squares tells us that $p\equiv 1 \bmod 4$.

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For the first part we have $p=a^2+b^2$. Take this modulo $a$ and it gives $p\equiv b^2\bmod a$. That is simply what modulo $a$ means.