Exercise 3.3.14* (If $p = a^2 + b^2$ with $a$ odd and positive, then $\genfrac{(}{)}{}{}{a}{p} = 1$)

Question

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Suppose that $p$ is prime, $p\equiv 1 \pmod 4$, and that $p = a^2 + b^2$ with $a$ odd and positive. Show that $\legendre{a}{p} = 1$.

Richard Ganaye · Accepted Answer

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} If $a = 1$, then $\legendre{a}{p} = 1$. Suppose now $a 
e 1$. Since $a>0$ and $a$ odd,  $a\geq 3$.

Write $a = p_1p_2\ldots p_l$ the decomposition of $a$ in prime factors, where the $p_i$ are odd primes, not necessarily distinct. Since $p = a^2 + b^2$, where $p_i \mid a$,
$$p \equiv b^2 \pmod {p_i}.$$
This shows that
$$\legendre{p}{p_i} = 1,\qquad i = 1,2,\ldots,l.$$
Therefore the Jacobi symbol $\legendre{p}{a}$ is
$$\legendre{p}{a} = \prod_{i=1}^l \legendre{p}{p_i} = 1.$$
By the law of quadratic reciprocity for the Jacobi symbol (Theorem 3.8), applied to the odd positive integers $a ,p$, where $p \equiv 1 \pmod 4$
$$\legendre{a}{p} = (-1)^{\frac{p-1}{2} \frac{a-1}{2}} \legendre{p}{a} = \legendre{p}{a} = 1.$$
So
$$\legendre{a}{p}  = 1.$$
\end{proof}

Exercise 3.3.14* (If $p = a^2 + b^2$ with $a$ odd and positive, then $\genfrac{(}{)}{}{}{a}{p} = 1$)

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Exercise 3.3.14* (If $p = a^2 + b^2$ with $a$ odd and positive, then $\genfrac{(}{)}{}{}{a}{p} = 1$)

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