This might have been one of the things that prompted the question, but I think Don Zagier’s one-sentence proof of Fermat’s theorem on sums of two squares ( https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares#Zagier's_"one-sentence_proof") uses this technique! Given a prime p = 4k+1, it defines a finite set S and two involutions. One involution’s fixed points (if they exist) are representations of p as a sum of two squares, while the other has exactly one fixed point. But replacing one involution with another either adds or removes fixed points in pairs, so the first involution has an odd and thus nonzero number of fixed points. (It may be too early for me to do math properly, but it seems like the second involution also shows that S has an odd number of elements, and so the first involution must have an odd number of fixed points?) (On a related note, in case anyone hasn’t seen it, https://mathoverflow.net/a/299696 has a nice geometric method of showing where the involution comes from!) --Neil Bickford On Wed, Jan 29, 2020 at 9:46 AM James Propp <jamespropp@gmail.com> wrote:
Sounds interesting. Do you have a reference?
Jim
On Wed, Jan 29, 2020 at 11:21 AM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
Maybe this is not quite what you want - but Another Hamiltoniain Path
says
that if there is a Hamiltonian path in G, then there is another one either in G or its complement \bar{G}.
The idea is to define an (exponentially large) graph in which Hamiltonian paths correspond to the number of odd-degree vertices. There are an even number of these.
So… this is sorta an example of odd parity, but it’s really showing that the set plus one has even parity :-)
Cris
On Jan 29, 2020, at 7:44 AM, James Propp <jamespropp@gmail.com> wrote:
What are people’s favorite examples of existence proofs that show that a set is not empty by showing that its cardinality is odd?
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