Why was this visual proof missed for 400 years? (Fermat's two square theorem)
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as a visually striking, accessible proof of the two square theorem, both in answers and as a supporting reference.
Recommended by 3 people across 3 different months, 2020 to 2023.
Why MathOverflow recommends it
MathOverflow users link it as the clearest explanation of what Zagier's terse one-sentence proof is actually doing.
Recommended by 1 people across 1 different months, 2020 to 2023.
What it is
A Mathologer video presenting a visual 'windmill' proof of Fermat's two square theorem (an odd prime is a sum of two squares iff it is 1 mod 4). It covers the theorem's history, Zagier's one-sentence proof, the windmill construction, and a proof of uniqueness of the representation.
Cited in 1 Wikipedia article, including Fermat's theorem on sums of two squares.
What people said
Strong rec Stephen Montgomery-Smith · 2020-08 · Mathematics Stack Exchange · on “Visually stunning math concepts which are easy to explain”
I just saw this movie on YouTube that gives a visual proof that every prime that is 1 modulo 4 can be written as the sum of two squares. This is a highly non-trivial theorem, and that it can be explained visually is impressive. Why was this visual proof missed for 400 years? (Fermat's two square theorem) [this video]
Recommends Todd Trimble · 2020-06 · MathOverflow · on “Request for examples: verifying vs understanding proofs”
I am one of those at MO who has complained about this proof. However, I was shown the light about what is really going on there by this pretty wonderful Mathologer video on youtube: [this video] (Oh, and now I see that Moritz Firsching explained the same thing in the other thread.)
Recommends Arthur · 2021-07 · Mathematics Stack Exchange · on “Primes congruent to $1$ mod $4$ are sum of two squares”
I always liked this proof [this video] . It is also featured in Proofs From The Book. It's not as straight-forward to invent as just passing to Gaussian integers, but it's so pretty.
Recommends user999691 · 2022-07 · Mathematics Stack Exchange · on “Fermat's Two Square Theorem: How do you prove that an odd prime is the sum of two squares iff it is congruent to 1 mod 4?”
See Geometrical figure explanation of Zagier's one-sentence proof: [LINK] See details explanation given By mathologer on his you tube video [this video]
Cites Timothy Chow · 2021-10 · MathOverflow · on “Reference request for a proof of the two-square Theorem”
As explained by Mathologer [this video] , Zagier's proof can be thought of as boiling down to showing that for primes $p \equiv 1\pmod{4}$, the equation $p = x^2 + 4yz$ has an odd number of positive integer solutions $(x,y,z)$. So it seems to be not quite the same proof. You might look at Christian Elsholtz's paper, A combinatorial approach to sums of two squares and related problems [LINK] , to see if your proof is implicitly in there somewhere.
Cites D.R. · 2023-11 · Mathematics Stack Exchange · on “Conceptual explanation for extra/missing $p$ solutions to $x^2+y^2=a \pmod p$ at $a=0$”
Throughout, $p$ will denote a prime integer, and $k$ an arbitrary integer. I have worked through V. Lebesgue's proof of quadratic reciprocity outlined by Keith Conrad in this MO thread [LINK] , and I feel that I have gained a conceptual understanding of it at all points except for the following "numerical coincidence": it is not difficult to see that the mod- $p$ norm map $\mathbb Z/p\mathbb Z[i] \to \mathbb Z/p\mathbb Z$ restricts to a (multiplicative) group homomorphism on units $(\mathbb Z/p\mathbb Z[i])^\times \to (\mathbb Z/p\mathbb Z)^\times$ and restricts to the $0$ -map outside the units; in particular we have that the fibers of the group homomorphism are all the same size ( $\implie
Cites D.R. · 2023-11 · MathOverflow · on “Conceptual explanation for extra/missing $p$ solutions to $x^2+y^2=a \pmod p$ at $a=0$”
Throughout, $p$ will denote a prime integer, and $k$ an arbitrary integer. I have worked through V. Lebesgue's proof of quadratic reciprocity outlined by Keith Conrad in this MO thread [LINK] , and I feel that I have gained a conceptual understanding of it at all points except for the following "numerical coincidence": it is not difficult to see that the mod- $p$ norm map $\mathbb Z/p\mathbb Z[i] \to \mathbb Z/p\mathbb Z$ restricts to a (multiplicative) group homomorphism on units $(\mathbb Z/p\mathbb Z[i])^\times \to (\mathbb Z/p\mathbb Z)^\times$ and restricts to the $0$ -map outside the units; in particular we have that the fibers of the group homomorphism are all the same size ( $\implie