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What is the content of Zagiers proof?

What is the actual proof and why does it work? I am not sure I understand why,

  1. there is only one fixed point, and

  2. why that implies that the involution $(x,y,z) \to (x,z,y)$ proves the theorem.

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3 Answers 3

up vote 5 down vote accepted
  1. It's unimportant that there's exactly one fixed point, what's important is that there's an odd number of them. Unfortunately, to see that it's an involution with a fixed point you need to do the calculation. It's not hard but it is a grind, which is why I feel it's a bit disingenuous to call this a "one sentence proof".

  2. Involutions either preserve points or swap a pair of them (as they are self-inverse), so if an odd number of points are preserved then the total order of the set is odd (if it is finite) because all other elements come in distinct pairs (those swapped by the involution). Hence the order of $S$ is odd. therefore all involutions of $S$ have a fixed point as if there were no fixed points then $|S|$ would be even.

  3. $(x,y,z) \mapsto (x,z,y)$ is an invoulution of $S$ so it has a fixed point, which implies that there's an $(x,y,z) \in S$ with $y = z$, hence $p = x^{2} + 4yz = x^{2} + 4y^{2} = x^{2} + (2y)^{2}$.

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Point of order: I'm new to this. Would it good practice for me to give the fixed point explicitly or better to let OP work it out? I'd guess the latter but am unsure. –  tharris Dec 24 '12 at 15:56
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Tom, there is no particular reason to avoid spelling things out for people, unless you have reason to believe the poster is doing homework or an exam, or merely asks for a hint... –  Thomas Andrews Dec 24 '12 at 16:16
    
Cheers. I thought maybe in this case it would be better not to say, as the proof leaves out the computational details as an exercise for the reader, so this is good practice for reading papers. Here of course those details amount to most of the work (but not the key idea). –  tharris Dec 24 '12 at 16:39
    
@Tom, after spending way more time than I should have, I understand the proof. Thank you very much for your help! –  picakhu Dec 24 '12 at 16:42

You have $x^2 + 4yz=p$. If the mapping $(x,y,z)\mapsto(x,z,y)$ has a fixed point, then that is a point where $(x,y,z)=(x,z,y)$. That implies $y=z$, so you've got $x^2 + 4y^2=p$. That implies $x^2+(2y)^2=p$, so $p$ is a sum of two squares.

(This is not a complete answer, but it answers at least one of the questions in the posting.)

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See this link and the references therein.

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While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  amWhy Dec 24 '12 at 15:41

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