[Guest]

If x, y, z all are positive integers... then

31x^3 - y^3 = z^3.

Now, if you consider non-positive integer solutions then..

(0, -1, 1) and (42, -65, 137) are some of the solutions...

So, find out positive solutions only

[Guest]

No solutions exisits for positive x,y and z

the complete proof is long and complicated so I will write only the main steps here:

31x³ = z³ + y³

31x³ = (z+y)(z²+y²-zy)

Then either z+y = 31n or z²+y²-zy = 31n

Consider first case where z+y = 31n

Then z²+y²-zy = (z+y)² - 3zy = 31²n² - 3zy

Let zy = an²

Then 31x³ = 31n(961n² - 3an²)

x³ = n³(961 - 3a)

Now 961 - 3a must be a cube; this is possible when "a" = 206 or 298 or 320

Since z+y = 31n for no value of "a" we can find integer solutions for z and y

For second case, let z²+y²-zy = 31n²

Then, z+y = n(31+3a)^1/2 where zy = an²

31x³ = n(31+3a)^1/2.31n²

This gives,

x³ = n³(31+3a)^1/2

x = n(31+3a)^1/6

Now in this case again, there is no value of a such that z and y can both be integers for an integer value of x

So there is no solution for x,y and z positive integers that satisfy this equation

Edited November 18, 2009 by DeeGee