Fermat's equation for odd exponents n asks for three integers, each with absolute value greater than 0, such that the sum of their nth powers is zero. A related problem is to find three integers, each with absolute value greater than the nth root of k, such that the sum of their nth powers equals k. For example, determine the integers x,y,z with |x|,|y|,|z| > 1 such that x^3 + y^3 + z^3 = 1 (1) This has infinitely many solutions because of the identity (1 +- 9m^3)^3 + (9m^4)^3 + (-9m^4 -+ 3m)^3 = 1 (2) but there are other solutions as well. Are there any other identities that give a different 1-parameter family of solutions? Is every solution of (1) a member of a family like this? In general it's known that there is no finite method for determining whether a given Diophantine equation has solutions. However, I wonder if there is a general method for determining if a given Diophantine equation has "algebraic" solutions, i.e., an algebraic identity like the one above that gives an infinite family of solutions. Does anyone know of such a method? (I suppose, in view of Falting's Theorem (formerly Mordell's Conjecture), that only equations of genus < 2 can have an algebraic solution.) It may be worth mentioning that the complete *rational* solution of the equation x^3 + y^3 + z^3 = t^3 is known, and is given by x = q [ 1 - (a - 3b)(a^2 + 3b^3) ] y = q [ (a + 3b)(a^2 + 3b^2) - 1 ] z = q [ (a^2 + 3b^2)^2 - (a + 3b) ] t = q [ (a^2 + 3b^2)^2 - (a - 3b) ] where q,a,b are any rational numbers. So if we set q equal to the inverse of [(a^2 + 3b^2)^2 - (a-3b)] we have rational solutions of (1). However, I think the problem of finding the _integer_ solutions is more difficult. If t is allowed to be any integer (not just 1) then Ramanujan gave the integer solutions x = 3n^2 + 5nm - 5m^2 y = 4n^2 - 4nm + 6m^2 z = 5n^2 - 5nm - 3m^2 t = 6n^2 - 4nm + 4m^2 This occassionally gives a solution of equation (1) (with appropriate changes in sign), as in the following cases n m x y z ---- ----- ------- ------- ------- 1 -1 (1) 2 -2 1 -2 9 10 -12 5 -12 -135 -138 172 19 -8 -791 -812 1010 46 -109 11161 11468 -14258 73 -173 65601 67402 -83802 419 -993 -951690 -926271 1183258 However, this doesn't cover all of the solutions given by (2). By the way, my "Most Wanted" problem #16 asks if the equation x^3 + y^3 + z^3 = 1 has any algebraic solutions other than (1 +- 9m^3)^3 + (9m^4)^3 + (-9m^4 -+ 3m)^3 = 1 and, if so, whether ALL the integer solutions are given by such an algebraic identity. Dean Hickerson (dean@Math.UCDavis.Edu) has informed me via email that there are known to be infinitely many algebraic solutions, and he cites the example (1 - 9 t^3 + 648 t^6 + 3888 t^9)^3 + (-135 t^4 + 3888 t^10)^3 + (3 t - 81 t^4 - 1296 t^7 - 3888 t^10)^3 = 1 However, he says it's not known whether EVERY solution of the equation lies in some family of solutions with an algebraic parameterization. He gives the following references "Sums of three cubes" by G. Payne and L. Vaserstein, in "The arithmetic of function fields: proceedings of the workshop at the Ohio State University, June 17-26, 1991", edited by David Goss, David R. Hayes, and Michael I. Rosen. "On the Diophantine equation x^3 + y^3 + z^3 = 1" by D. H. Lehmer, J. London Math. Soc. 31 (1956), 275-280. Interestingly, Dean notes that if you replace 1 by 2, then again there's a parametric solution: (6 t^3 + 1)^3 - (6 t^3 - 1)^3 - (6 t^2)^3 = 2 but again this doesn't cover all known integer solutions, such as 1214928^3 + 3480205^3 - 3528875^3 = 2 It's evidently not known if there are ANY other algebraic solutions besides the one noted above. In general it seems to be a difficult problem to characterize all the solutions of x^3 + y^3 + z^3 = k for some arbitrary integer k. In particular, the question of whether ALL integer solutions are given by an algebraic identity seems both difficult and interesting.

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