A challenge inspired from a question of Doron Zeilberger. Do there exist arbitrarily large integers \(n\) with the following property:
- There exists an ordered field \(F\) such that \(x^n+ y^n = z^n\) has solutions in \(F\) with \(xyz \ne 0\).
- The only solutions in \(F\) to \(x^m + y^m = z^m\) for \(3 \le m < n\) satisfy \(xyz = 0\),
To give a somewhat looser phrasing, you might try to prove Fermat over \(\mathbf{Q}\) by an inductive argument that only relies on positivity of squares together with the fact that Fermat was classical known for some small values of \(n\). This question asks whether you can rule out such a proof.
This might be tricky. Quite possibly taking \(F = \mathbf{Q}(2^{1/n}) \subset \mathbf{R}\) will work for infinitely many integers \(n\), but this is not obvious. Indeed, since any ordered field \(F\) will always contain \(\mathbf{Q}\), any proof that arbitrarily large \(n\) with the properties above exist will also prove Fermat over \(\mathbf{Q}\). That said, there might be simple constructions of such \(F\) assuming Fermat is true over \(\mathbf{Q}\) which we fortunately know to be true.
You want the strict inequality m n.
We can try to prove that for some t, t < 1, t rational, the Fermat curve of degree m, X_m, has no rational point over Q(b), b^n = 1 – t^n. Such a point would give rise to an effective divisor of degree n on X_m defined over the cyclotomic field K of n-th roots of unity (this field does not depend on t!). Our assumption is that the genus of X_m is bigger than n so the variety parametrizing effective divisors of degree n on X_m is a subvariety of the Jacobian of X_m to which we can apply Mordell-Lang (proved by Faltings) and hopefully (this I haven't checked) show that it can't have all these points that we get by varying t. We would have to make the argument generate enough t's to deal with the different m.
part of my post didn’t show up. I am assuming (m-1)(m-2)/2 > n.
Should the condition be m < n in the second property?
Fixed!
Clearly this can only work for n prime or 4, because of the usual argument reducing Fermat to those cases.
A variant of Felipe’s strategy is to take F = Q ( a,b)/[ a^n + b^n-1] with some arbitrary ordering. The advantage here is that, together with Fermat over Q, to check that there is no solution of the degree m equation reduces to a geometric question – there is no nonconstant map from the n’th Fermat curve to the m’the Fermat curve defined over Q (so, not purely geometric).
For m<n and n a prime p, a good approach might be to note that ell-adic Tate module of the p'th Fermat curve in unramified at primes away from p, but the Tate module of the m'th Fermat curve is ramified at primes dividing m, and a nonconstant map of curves is a surjection of Tate modules.
So indeed I think there is a simple construction assuming Fermat is true over Q.
To prove there is no map, you can also try the following. Take a prime p with p=-1 mod n and p = 1 mod m and work in F_p. The Fermat curve of degree n is supersingular and the one of degree m is ordinary (I think?).