Exercise 9.31

Proof. Recall the following lemma :

Lemma :

Let $p$be prime, $p\equiv 1[4]$, then ${\left[\left(\frac{p-1}{2}\right)!\right]}^2\equiv -1[p]$.

By Wilson’s theorem (Prop. 4.1.1, Corollary), $(p-1)!\equiv -1[p]$.

$\text{∙}$

We show that there exists a pair $a,b\in \mathbb{Z}$ which verifies the sentence.

By lemma 5 section 7, as $p\equiv 1[4]$, there exists an irreducible $\pi$ such that $N(\pi )=p$, and we can choose $\pi$ such that $\pi =A+\mathit{Bi}$ is primary (lemma 7 section 7), so $A$ is odd.

If $A\equiv 1(\mathit{mod}4)$, we take $a=A$, and if $A\equiv 3(\mathit{mod}4)$, we take $a=-A$ : then $a\equiv 1(\mathit{mod}4)$.

Let $u=\left(\frac{p-1}{2}\right)!$. Then $0\equiv p=A^2+B^2(\mathit{mod}p)$, $B^2\equiv -A^2\equiv {(\mathit{uA})}^2(\mathit{mod}p)$.

$p\mid (B-\mathit{uA})(B+\mathit{uA})$, thus $B\equiv \pm \mathit{uA}(\mathit{mod}p)$.

Since $a=\pm A$, $B\equiv \pm \mathit{ua}(\mathit{mod}p)$.

If $B\equiv -\mathit{ua}(\mathit{mod}p)$, we take $b=B$, if not $b=-B$.

Then $a,b$ are such that $p=a^2+b^2,a\equiv 1[4],b\equiv -((p-1)∕2)!a[p]$.

$\text{∙}$

Unicity of the pair $(a,b)$ such that $$p=a^2+b^2,a\equiv 1[4],b\equiv -((p-1)∕2)!a[p].$$

Suppose that $c,d$ are such that $p=c^2+d^2,c\equiv 1[4],d\equiv -((p-1)∕2)!c[p]$.

Let $\pi =a+\mathit{ib},\lambda =c+\mathit{id}$. As $p=\mathit{N\pi }=\mathit{N\lambda }$ is a rational prime, $\pi$ and $\lambda$ are primes in $D$, and $p=\pi \bar{\pi }=\lambda \bar{\lambda }$, thus $\lambda$ is associate to $\pi$ or $\bar{\pi }$. :

$$\lambda \in \{\pi ,-\pi ,\mathit{i\pi },-\mathit{i\pi },\bar{\pi },-\bar{\pi },i\bar{\pi },-i\bar{\pi }\}.$$

As $a,c$ are odd, and $b,d$ even, it remains only the possibilities $\lambda =\pm \pi ,\lambda =\pm \bar{\pi }$, thus $c=\pm a$. Moreover $a\equiv c\equiv 1[4]$, thus $a=c$, and $d\equiv -((p-1)∕2)!c\equiv -((p-1)∕2)!a\equiv b[p]$.

$p=a^2+b^2=a^2+d^2$, so $d=\pm b$, and $d\equiv b[p]$.

If $d=-b$, then $p\mid 2b$, thus $p\mid b$, and also $p\mid a$, so $p^2\mid a^2+b^2=p$: this is impossible. So $a=c,b=d$. Unicity is proved.

Conclusion : if $p\equiv 1[4]$, there exists an unique pair $a,b$ such that

$$p=a^2+b^2,a\equiv 1(\mathit{mod}4),b\equiv -((p-1)∕2)!a(modp).$$