Answer to Question #110560 in Combinatorics | Number Theory for Amy Wang

Question

Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?

Answer 1

To satisfy equation $x^3+y^3+z^3=42$ , the integers should be the following:

$(-80\ 538\ 738\ 812\ 075\ 974)^3 + 80\ 435\ 758\ 145\ 817\ 515^3 + 12\ 602\ 123\ 297\ 335\ 631^3 = 42$ .

It was figured out after 1.3 million hours of computing on the Charity Engine global grid (see https://en.wikipedia.org/wiki/Sums_of_three_cubes#Computational_results)