Exercise 5.26

Proof. Let $p$ a prime number, $p\equiv 1[4]$ : $p=4k+1,k\in {\mathbb{N}}^{\text{∗}}$.

Then $p=a^2+b^2$ (Ex. 5.24).

As $a,b$ are not of the same parity, up to exchange $a$ and $b$, we will suppose that $a$ is odd (then $b$ is even).

(a)

Using the law of quadratic reciprocity for Jacobi’s symbol (Proposition 5.2.2), where $a,p$ are odd numbers :

since $p\equiv 1[4]$.

If $a=p_1{p}_2\cdots p_l$ is the decomposition of $a$ in prime factors, with not necessarily distinct primes , then

Since $p=a^2+b^2$, $p\equiv b^2[p_i]$, thus $\left(\frac{p}{p_i}\right)=1$ for all $i$.

(b) $a+b$ is odd, and $p\equiv 1[4]$, thus

If $a+b=q_1{q}_2\cdots q_l$, as $2p={(a+b)}^2+{(a-b)}^2$, $2p\equiv {(a-b)}^2[q_i]$, thus $\left(\frac{2p}{q_i}\right)=1$.

Moreover $\left(\frac{2}{a+b}\right)={(-1)}^{\frac{{(a+b)}^2-1}{8}}$, so

(c) ${(a+b)}^2=a^2+b^2+2\mathit{ab}=p+2\mathit{ab}\equiv 2\mathit{ab}[p]$

(d) ${[{(a+b)}^2]}^{\frac{p-1}{4}}\equiv {(2\mathit{ab})}^{\frac{p-1}{4}}[p]$, thus

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