This was posted to Math Overflow in 2018, but I just found it via a comment on @Mathologer’s video https://www.youtube.com/watch?v=DjI1NICfjOk …, which also explains this proof.
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The idea, briefly, is: The transformation illustrated transforms every windmill to a different windmill of the same area; cross-shaped windmills (whose arms are the same width as the central square) are mapped to themselves.
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There is at most one cross-shaped windmill with a given prime area, since the width of the arms must divide the total area (of a cross-shaped windmill). If p = 4k + 1 then there exists a cross-shaped windmill of area p (with four arms of width 1 and length k).
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So… if p = 4k + 1 is prime, there are an *odd* number of windmills with area p.
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Now consider the transformation that rotates the windmill arms by 90°. This doesn’t change the area either. Since there are an odd number of windmills of area p, the arm-rotation transformation must also have a fixed point, i.e. there must be a windmill with square arms.
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So p is the sum of a square and four times a square – but four times a square is still a square!
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It is easier to see now why its name is a "Pythagorean tessagon" https://github.com/cwant/tessagon/blob/master/documentation/pythagorean_tessagon.md …pic.twitter.com/gSjAzh68iU
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Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
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Oh, fabulous! For years I've had in my list of things to do "Understand Zagier's proof" but never found time. This is lovely! I already give a proof of this as part of maths workshops and maths club talks, this will be a brilliant kicker. FWIW, the proof I give is this ...
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Suppose p=1 (4). Then by Wilson's theorem, (p-1)! = -1 (p). We leverage that to show that ((p-1)/2)! is a square root of -1 (mod p). That *really* get maths club kids excited. Then let a and b be between 0 and sqrt(p) inclusive. There are ceil(sqrt(p)) values for each. /n
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