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.

Congratulations! This is very nice work…

[...] 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 [...]

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.