4 Representing numbers as sums of two squares
For primes \(p = 4m + 1\) the equation \(s^2 \equiv -1 (\mod p)\) has two solutions \(s \in \{ 1, 2, \dots , p - 1\} \), for \(p = 2\) there is one such solution, while for primes of the form \(p = 4m + 3\) there is no solution.
TODO
No number \(n = 4m + 3\) is a sum of two squares.
TODO
Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).
TODO
Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).
TODO (Zagier’s one line proof is in mathlib by now, follow this!)
Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).
TODO
A natural number \(n\) can be represented as a sum of two squares if and only if every prime factor of the form \(p = 4m + 3\) appears with an even exponent in the prime decomposition of \(n\).
TODO