A^4 + B^2 = C^n

Mike Bennett, Nathan Ng, and I posted a paper recently about solutions to the equation

A^4 + B^2 = C^n (*)

in relatively prime integers (A,B,C). There are infinitely many solutions to this equation when n = 2 or 3; for example,

1089^4 + 1549034^2 = 15613^3.

By contrast, what Mike, Nathan and I prove is that for n > 3, there are no solutions to (*).

(CORRECTION:  Mrs. Q points out that I need to say “relatively prime nonzero integers” above — otherwise there are solutions like 0^4 + 1^2 = 1^n.)

More math below the fold:

This problem is part of a class of problems known as ternary Diophantine equations or generalized Fermat equations: one asks whether

A^p + B^q = C^r (**)

has any solution in relatively prime integers (A,B,C).

(Exercise: show that if we don’t impose the condition that (A,B,C) be relatively prime, there are always infinitely many solutions to (**), so it’s not such an interesting question.)

We denote by GF(p,q,r) the problem of finding all solutions to (**). The most famous case, of course, is that where p=q=r! This is the classical conjecture of Fermat, finally resolved by Wiles and Taylor-Wiles in 1995 after a few hundred years of hard work by many hands.

Theorem: (Wiles, Taylor-Wiles, 1995) GF(p,p,p) has no solutions for p > 2.

You might ask: what if it’s not possible to write down all solutions to GF(p,q,r) — what if, for instance, there are infinitely many? A beautiful theorem of Henri Darmon and Andrew Granville demonstrates that, apart from very small values of p,q,r, this isn’t possible.

Theorem: (Darmon-Granville, 1995) Suppose that 1/p + 1/q + 1/r < 1. Then GF(p,q,r) has only finitely many solutions.

The proof reduces the problem to the famous theorem of Faltings, that a curve of genus at least 2 has only finitely many rational points. But there’s actually a very nice heuristic argument which (disadvantage) doesn’t prove anything, but (advantage) makes it clear why this funny condition 1/p + 1/q + 1/r < 2 is relevant.

Heuristic: In the interval [0..X] there are about X^(1/p) pth powers, and about X^(1/q) qth powers. So the number of sums of the form A^p + B^q is about X^(1/p + 1/q). Now what’s the chance that you get incredibly lucky and this sum is actually a perfect rth power? Well, the sum is also going to be in [0..X] (more precisely [0..2X], but we won’t trouble ourselves with constants!) There are X^(1/r) rth powers in this interval, so the chance that any particular number is an rth power is X^(1-1/r). So the expected number of solutions to (**) in this interval is

X^(1/p + 1/q + 1/r – 1)

and what the Darmon-Granville condition says is precisely that the exponent is negative — that in a large interval we should expect to see approximately zero solutions.

In general, very little is known about GF(p,q,r). A very incomplete sampling of existing results: the cases of GF(p,p,2) and GF(p,p,3) were handled by Darmon and Loic Merel in 1997; (3,3,p) by Alain Kraus in 1998; (2,3,8) and (2,3,9) by Nils Bruin in 2000 and 2004; and (2,3,7), in a technical tour de force by Bjorn Poonen, Ed Schaefer, and Michael Stoll, just last year. The only solution to this last equation (in positive integers) is

15312283^2 + 9262^3 = 113^7.

Now things get a bit more technical. A few years ago, I proved that GF(2,4,p) had no solutions for p > 211. The proof follows the general lines of the Wiles argument, using theorems about modularity of elliptic curves. However, I had to use a step from analytic number theory — namely, I needed to estimate the average special value of a family of L-functions, using an idea of Bill Duke. The fact that I could only estimate this average, not compute it exactly, led to the fact that I could only handle GF(2,4,p) for fairly large values of p. So a few times a year people would ask me, “What about smaller values of p? Can they be handled?” Now I can say yes — between us, we were able to bring to bear enough techniques to settle the problem completely. What goes into it: a sharpening of my original estimate, some applications of Chabauty techniques a la Bruin, elimination of various p by arguments on Galois representations, and finally direct computation of special values in MAGMA to take care of the remaining cases. The result is a bit of a patchwork, and I don’t know if it’ll be fun to read — but it was fun to do!

3 thoughts on “A^4 + B^2 = C^n”

1. Congratulations! This is very nice work…

2. […] serious math. Quomodocumque (Jordan Ellenberg) publishes an accessible summary of a recent paper on the equation and its solutions. A bit reminiscent of Fermat’s Last Theorem? Yup. 02 – Charles Daney asks “Why should […]

3. KCd says:

You write that the only solution to x^2 + y^3 = z^7 in positive relatively prime integers is 15312283^2 + 9262^3 = 113^7. There is also 2213459^2 + 1414^3 = 65^7.