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82 lines
4.3 KiB
Markdown
82 lines
4.3 KiB
Markdown
---
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lastmod: "2021-09-12T22:47:44.0000000+01:00"
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author: patrick
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categories:
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- mathematical_summary
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comments: true
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date: "2014-09-09T00:00:00Z"
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math: true
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aliases:
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- /mathematical_summary/sum-of-two-squares-theorem/
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- /sum-of-two-squares-theorem/
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title: Sum-of-two-squares theorem
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---
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*Wherein I detail the most beautiful proof of a theorem I've ever seen, in a bite-size form suitable for an Anki deck.
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# Statement
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There's no particularly nice way to motivate this in this context, I'm afraid, so we'll just dive in. I have found this method extremely hard to motivate - a few of the steps are a glorious magic.
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* \\(n\\) is a sum of two squares iff in the prime factorisation of \\(n\\), primes 3 mod 4 appear only to even powers.
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# Proof
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We're going to need a few background results.
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## Background
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* \\(\mathbb{Z}[i]\\), the ring of [Gaussian integers], is a UFD.
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* In a UFD, [irreducible]s are [prime].
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* \\(-1\\) is square mod \\(p\\) iff \\(p\\) is not 3 mod 4.
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Additionally, we'll call a number which is the sum of two squares a **nice** number.
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## First implication: if primes 3 mod 4 appear only to even powers…
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We prove the result first for the primes, and will then show that niceness is preserved on taking products.
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* Let \\(p=2\\). Then \\(p\\) is trivially the sum of two squares: it is \\(1+1\\).
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* Let \\(p\\) be 1 mod 4.
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* Then modulo \\(p\\), we have \\(-1\\) is square.
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* That is, there is \\(n \in \mathbb{N}\\) such that \\(x^2 + 1 = n p\\).
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* That is, there is \\(n \in \mathbb{N}\\) such that \\((x+i)(x-i) = n p\\).
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* \\(p\\) divides \\((x+i)(x-i)\\), but it does not divide either of the two multiplicands (since it does not divide their imaginary parts).
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* Therefore \\(p\\) is not prime in the complex integers.
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* Since \\(\mathbb{Z}[i]\\) is a UFD, \\(p\\) is not irreducible in the complex integers.
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* Hence there exist non-invertible \\(a, b \in \mathbb{Z}[i]\\) such that \\(a b = p\\).
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* Taking norms, \\(N(p) = N(ab)\\).
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* Since the norm is multiplicative, \\(N(p) = N(a) N(b)\\).
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* \\(N(p) = p^2\\), so \\(p^2 = N(a) N(b)\\).
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* Neither \\(a\\) nor \\(b\\) was invertible, so neither of them has norm 1 (since in \\(Z[i]\\), having norm 1 is equivalent to being invertible).
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* Hence wlog \\(N(a)\\) is exactly \\(p\\), since the product of two numbers being \\(p^2\\) means either one of them is 1 or they are both \\(p\\).
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* Let \\(a = u+iv\\). Then \\(N(a) = u^2 + v^2 = p\\), which was what we needed.
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Next, we need to take care of this "even powers" business:
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* \\(p^2\\) is a sum of two squares if \\(p\\) is 3 mod 4: indeed, it is \\(0 + p^2\\).
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All we now need is for niceness to be preserved under multiplication. (Recall \\(w^*\\) denotes the conjugate of \\(w\\).)
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* Let \\(x, y\\) be the sum of two squares each, \\(x_1^2 + x_2^2\\) and \\(y_1^2 + y_2^2\\).
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* Then \\(x = (x_1 + i x_2)(x_1 - i x_2)\\), and similarly for \\(y\\).
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* Then \\(x y = (x_1 + i x_2)(x_1 - i x_2)(y_1 + i y_2)(y_1 - i y_2)\\).
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* So \\(x y = w w^*\\), where \\(w = (x_1 + i x_2)(y_1 + i y_2)\\).
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* Hence \\(x y = N(w)\\), so is a sum of two squares (since norms are precisely sums of two squares).
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Together, this is enough to prove the first direction of the theorem.
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## Second implication: if \\(n\\) is the sum of two squares…
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We'll suppose that \\(n = x^2 + y^2\\) has a prime factor which is 3 mod 4, and show that it divides both \\(x\\) and \\(y\\).
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* Let \\(n = x^2 + y^2\\) have prime factor \\(p\\) which is 3 mod 4.
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* Then taken mod \\(p\\), we have \\(x^2 + y^2 = 0\\).
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* That is, \\(x^2 = - y^2\\).
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* If \\(y\\) is not zero mod \\(p\\), it is invertible.
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* That is, \\((x y^{-1})^2 = -1\\).
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* This contradicts that \\(p\\) is 3 mod 4 (since \\(-1\\) is not square mod \\(p\\)). So \\(y\\) is divisible by \\(p\\).
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* Symmetrically, \\(x\\) is divisible by \\(p\\).
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* Hence \\(p^2\\) divides \\(n\\), so we can divide through by it and repeat inductively.
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That ends the proof. Its beauty lies in the way it regards sums of two squares as norms of complex integers, and dances into and out of \\(\mathbb{C}\\), \\(\mathbb{Z}[i]\\) and \\(\mathbb{Z}\\) where necessary.
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[Gaussian integers]: https://en.wikipedia.org/wiki/Gaussian_integers
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[UFD]: https://en.wikipedia.org/wiki/Unique_factorization_domain
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[irreducible]: https://en.wikipedia.org/wiki/Irreducible_element
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[prime]: https://en.wikipedia.org/wiki/Prime_element |