Zagier's one-sentence proof of Fermat's theorem. - MathOverflow

Zagier's one-sentence proof of Fermat's theorem.

2010-07-08T20:36:53Z

Zagier has a very short proof ( MR1041893) for the fact that every prime number $p$ of the form $4k+1$ is the sum of two squares. The proof defines an involution of the set $S= \lbrace (x,y,z) \in N^3: x^2+4yz=p \rbrace $ which is easily seen to have exactly one fixed point. This shows that the involution that swaps $y$ and$ z$ has a fixed point too, implying the theorem.

This simple proof has always been quite mysterious to me. Looking at a precursor of this proof by Heath-Brown did not make it easier to see what, if anything, is going behind the scene. There are similar proofs for the representation of primes using some other quadratic forms, with much more involved involutions.

Now, my question is: is there any way to see where these involutions come from and to what extent they can be used to prove similar statements?

Answer by Daniel Litt for Zagier's one-sentence proof of Fermat's theorem.

2010-07-08T21:25:43Z

This paper by Christian Elsholtz seems to be exactly what you're looking for. It motivates the Zagier/Liouville/Heath-Brown proof and uses the method to prove some other similar statements. Here is a German version, with slightly different content.

Essentially, Elsholtz takes the idea of using a group action and examining orbits as given (and why not -- it's relatively common) and writes down the axioms such a group action would have to fulfill to be useful in a proof of the two-squares theorem. He then algorithmically determines that there is a unique group action satisfying his axioms -- that is, the one in the Zagier proof. The important thing is that having written down these (fairly natural) axioms, there's no cleverness required; finding the involution in Zagier's proof boils down to solving a system of equations.

Answer by Franz Lemmermeyer for Zagier's one-sentence proof of Fermat's theorem.

2010-07-10T18:48:11Z

It's been a while since I read Elsholtz's article, but after doing so I felt none the wiser. Below I have translated Heath-Brown's proof into the language of binary quadratic forms; Zagier's proof looks more interesting from this point of view (the connections to Gauss reduction are much closer), but when working out the details I got stuck in the middle.

One essential ingredient for the proofs by Heath-Brown and Zagier was pointed out already by Frick in 1918, who showed that if $p = a^2 + 4b^2$ is an odd prime number, then the indefinite binary quadratic form $Q = (-b,a,b)$ with discriminant $p$ is Gauss reduced and is contained in the principal cycle.

For proving that such a form exists without assuming that $p$ is a sum of two squares, we consider all forms $(A,B,C)$ with discriminant $p$ such that $A < 0$ and $C > 0$. From $p = B^2 - 4AC$ it then follows that the set $$ S = {(A,B,C): B^2 - 4AC = p, A < 0, C > 0} $$ is finite. The obvious map $$ \mu: S \to S, \quad (A,B,C) \to (-C,B,-A) $$ is an involution; if $S$ had odd cardinality, it would follow that $\mu$ has a fixed point, say $(A,B,-A)$, from which we would get $p = B^2 + 4A^2$. Unfortunately, $S$ has even cardinality since the involution $$ \nu: S \to S, \quad (A,B,C) \to (A,-B,C) $$ has no fixed points: this is because $B = 0$ implies $p = 4AC$, which is impossible for prime numbers $p$.

We now would like to find a subset $U \subset S$ of $S$ with odd cardinality on which $\mu$ is still defined. The most natural idea would be considering the forms with $B > 0$. For showing that this set of forms has odd cardinality, we have to define an involution $(A,B,C) \to (A',B',C')$ on this subset that has exactly one fixed point. To find such an involution, we start with $(A,B,C) \to (A,-B,C)$ and then apply reduction by changing the middle coefficient modulo $2A$ and then adjusting the last coefficient so that the discriminant is $p$. This gives $$ (A,-B,C) \to (A',B',C') = (A,-2A-B,A+B+C). $$ Now we are facing the problem that it is not clear at all that $B' = -2A-B > 0$, or that $C' = A+B+C > 0$. But if we set $$ U = {(A,B,C) \in S: A+B+C > 0 }, $$ then the map $$ \gamma: (A,B,C) \to (A,-2A-B,A+B+C) $$ actually is an involution on $U$. Moreover, $(A,B,C)$ is a fixed point if and only if $-2A-B = B$ and $A+B+C = C$, which is equivalent to $A = -B$. Since $p = B^2 - 4AC = B^2 + 4BC = B(B+4C)$ is prime, we must have $|A| = |B| = 1$. Since $A < 0$, this implies that the fixed point is $(-1,1,\frac{p-1}4)$; this form is equivalent to the principal form $(1,1,\frac{p-1}4)$.

The involution $\gamma$ on $U$ shows that $U$ has odd cardinality; the map $$ (A,B,C) \to (-C,-B,-A) $$ is an involution on $S$ sending $U$ to $S \setminus U$, which impliesthat $|S| = 2 |U|$. The involution $\nu$ on $S$ sends elements with $B > 0$ to elements with $B < 0$, hence $$ T = {(A,B,C) \in S: B > 0} $$ has the same number of elements as $U$, and in particular, it has odd cardinality. Finally, $\mu$ is an involution on $T$, and now the Two-Squares Theorem follows.

References

H. Frick, Über den Zusammenhang der Perioden quadratischer Formen positiver Determinante mit der Zerlegung einer Zahl in die Summe zweier Quadrate, Diss. ETH Zürich, 1918

Answer by Christian Elsholtz for Zagier's one-sentence proof of Fermat's theorem.

2010-07-16T14:28:27Z

As the answers above linked to an old paper of mine (in German, and a somewhat different English preprint), some readers might like to know that an updated version is to appear very soon and is now linked on my webpage:

http://www.math.tugraz.at/~elsholtz/WWW/papers/papers30nathanson-new-address3.pdf

In addition to the motivation of the Heath-Brown/Zagier proof it contains for example

a) a discussion of a lattice point proof (section 1.6)

b) much more historical information and links to other work

c) an alternative motivation of the Heath-Brown-Zagier proof, due to Dijkstra (section 2.3)