Numbers of the form $x^2 + y^2$

Question

Which numbers can be represented in the form of $x^2 + y^2$, $x,y \in\mathbb N$?

My approach : I found out that for this to satisfy, one of the prime factors of the number has to be congruent to $1$ (mod $4$) . Is there any other elegant method for this

See https://wstein.org/edu/Fall2001/124/lectures/lecture21/lecture21/node2.html — Mark Saving, Jul 16 '21 at 18:15
This is essentially equivalent to figuring out the behavior of primes in $\mathbb Z[\sqrt{-1}]$. — Kenta S, Jul 16 '21 at 18:26

score 2 · Accepted Answer · answered Jul 16 '21 at 18:20

Let $A$ be the set of all prime numbers of the form $4k+3$. Then, $n$ is the sum of two squares if and only if $\forall p \in A\, p \mid n \implies 2 \mid \nu_p(n)$, where $\nu_p(n)$ is the $p$-adic valuation of $n$. In other words, $\nu_p(n)$ must be an even number for all primes $p$ of the form $4k+3$ dividing $n$.

score 1 · Answer 2 · answered Jul 16 '21 at 18:15

1

Integers that in their prime factorization, contain either only odd primes of the form $4k+1$ or if there is a prime factor of the form $4k+3$, then the multiplicity of this prime should be even, can be written as the sum of two squares.

answered Jul 16 '21 at 18:15

F. A. Mala

1

score 1 · Answer 3 · answered Jul 16 '21 at 19:19

I will outline a proof of the characterization already given by Geoffrey Trang.

Step 1. The primes of the form $x^2+y^2$ are $2$ and the primes $\equiv 1\pmod{4}$.
Trivially, a prime $\equiv 3\pmod{4}$ cannot be represented by $x^2+y^2$ and $2=1^2+1^2$.
If $p\equiv 1\pmod{4}$ the Legendre symbol $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}$ gives that $-1$ is a quadratic residue in $\mathbb{F}_p$, so for some integer $a\leq\frac{p-1}{2}$ we have $a^2+1= kp$ with $k<\frac{p}{4}$. The fact that the norm in $\mathbb{Z}[i]$ is multiplicative leads to Lagrange's identity $$ (a^2+b^2)(c^2+d^2) = (ad-bc)^2+(ac+bd)^2. $$ By Fermat's descent we are able to go from $a^2+1=kp$ to $a^2+b^2=p$: every prime $\equiv 1\pmod{4}$ can be written as a sum of two squares in a essentially unique way. By Lagrange's identity once again we have that any $n$ without prime factors of the form $4k+3$ can be written as a sum of two squares. The same holds if for every prime divisor $p$ of the form $4k+3$ we have $\nu_p(n)\equiv 0\pmod{2}$.

Step 2. If there is a prime divisor $p$ of the form $4k+3$, such that $\nu_p(n)$ is odd, $n$ cannot be written as a sum of two squares. In $\mathbb{F}_p$ we have that $-1$ is not a quadratic residue, hence $x^2+y^2\equiv 0\pmod{p}$ implies that both $x$ and $y$ are multiples of $p$. In turn this implies that $\nu_p(n)$ is even, contradiction.

Note. $21\equiv 1\pmod{4}$ but it cannot be written as a sum of two squares, since $\nu_3(21)$ is odd.
Something similar applies to $2021=43\cdot 47$.

Numbers of the form $x^2 + y^2$

3 Answers3