Fermat-like Equations (noch nicht übersetzt)

Problem 678

If a triple of positive integers (a,b,c) satisfies a2+b2=c2, it is called a Pythagorean triple. No triple (a,b,c) satisfies ae+be=ce when e3 (Fermat's Last Theorem). However, if the exponents of the left-hand side and right-hand side differ, this is not true. For example, 33+63=35.

Let a,b,c,e,f be all positive integers, 0<a<b, e2, f3 and cfN. Let F(N) be the number of (a,b,c,e,f) such that ae+be=cf. You are given F(103)=7, F(105)=53 and F(107)=287.

Find F(1018).