this infinite family of solutions puts the kibosh on the way i like to think about equations like x^3 + y^3 = z^3: namely, are the cubes dense enough that (set of cubes) + (set of cubes) intersects (set of cubes) ? in a sense, flt says no. but, by adding 1 to each element on the right-hand side, you get an infinite set of examples where (set of cubes) + (set of cubes) intersects (1+set of cubes). weird. bob baillie --- Fred W. Helenius wrote:
At 07:08 AM 7/22/2008, rwg@sdf.lonestar.org wrote:
New business: Can someone tell me if Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124,
or some other source answers whether A050788 is known infinite, i.e., does c^3+1=a^3+b^3 infinitely often?
There is an infinite family of solutions given by (a,b,c) = (9n^3 + 1, 9n^4, 9n^4 + 3n). A050788 actually asks about x^3+y^3=z^3-1 with x < y < z; for that we can take (x,y,z) = (9n^3 - 1, 9n^4 - 3n, 9n^4) for n > 1.
I extracted these solutions from Theorem 235 in Hardy & Wright; the result shown there is that all nontrivial rational solutions of x^3 + y^3 = u^3 + v^3 are given by
x = r(1 - (a - 3b)(a^2 + 3b^2)) y = r((a + 3b)(a^2 + 3b^2) - 1) u = r((a + 3b) - (a^2 + 3b^2)^2) v = r((a^2 + 3b^2)^2 - (a - 3b))
where r,a,b are rational and r is not zero.
Specializing to r = 1, b = n/2 and a = 3n/2 gives
x = 1 y = 9n^3 - 1 u = 3n - 9n^4 v = 9n^4.
The solutions given above are obtained by changing signs and moving cubes from one side of the equation to the other as necessary. Unfortunately, not all integral solutions are found so easily: the third value in A050788 corresponds to 135^3 + 138^3 = 172^3 - 1; this is not produced by such simple choices of r,a,b.
And does ceiling((a^n+b^n)^(1/n))^n - a^n - b^n < k have only finitely many solutions for n>3?
With a,b positive that would be my guess, but I don't expect a proof any time soon.