$p=a^2+b^2,$ $p\equiv 1\pmod 4$, solve for $(a,b)$

Question

Let $p\equiv 1\pmod 4$ be prime. Then $p$ splits in $\mathbb{Z}[i]$ into non-units $(a+ib)(a-ib)=p$. Thus it is clear that $p=a^2+b^2$.

Question: How does one determine $(a,b)$?

According to Dedekind-Kummer (the extension is monogenic) and with $\mathfrak{p}=(p)$ in $\mathbb{Z}$, we have that $\mathfrak{p}\mathbb{Z}[i]=(p,\zeta_4-x)(p,\zeta_4-y)$, where $\zeta_4$ is a primitive $4$-th root of unity, and $x,y$ are the two primitive $4$-th roots of unity modulo $p$.

But $\mathbb{Z}[i]$ is a PID, so can write these ideals with a principal element, which should be $a+ib$ and $a-ib$. How can one do this?

Note that $\mathbb Z[i]$ is even a Euclidean domain. Hence, in concrete situations, you can apply the Euclidean algorithm to, say, $p$ and $\zeta_4-x$ to obtain their gcd. — Claudius, Dec 19 '21 at 21:47

score 2 · Answer 1 · answered Dec 19 '21 at 22:22

2

Find integer $x>0$ such that $p$ divides $x^2+1$.
Find $u,v$ between $1$ and $\lfloor \sqrt{p}\rfloor$ with $xu\equiv \pm v\pmod{p}$.
Then $x^2+1\equiv 0\pmod{p}$ implies $x^2u^2+u^2\equiv v^2+u^2\equiv 0\pmod{p}$.
But $v^2+u^2<2p$, so $v^2+u^2=p$.

answered Dec 19 '21 at 22:22

markvs

19,653

The Euclidean algorithm mentioned by Claudius seems to be faster than this. – Tejas Rao Dec 19 '21 at 22:37
The Euclidean algorithm may be faster, but I did not check all details. As for "better", it depends on the goal. – markvs Dec 19 '21 at 22:41

$p=a^2+b^2,$ $p\equiv 1\pmod 4$, solve for $(a,b)$

1 Answers1