He he. The reason it is not easy is that the exponents are odd. This allows the sign to carry through and essentially allows the domain of solutions to be infinite. Conversely, even powered versions, such as

x^(2k) + y^(2k) + z^(2k) = N, k is a positive integer, are much easier, because

x^(2k) = N - y^(2k) - z^(2k). Since y^(2k) and z^(2k) must be >= 0, then the maximum value that any one term can have is where the other two are zero, therefore a loose bound to the number space is

x^(2k) = N .... x = floor(N^(1/2k)). This is symmetric. (Ignoring the depression that occurs when other terms are non-zero. (The extreme case is when they are all equal... 3*x^(2k) = N .... x = floor((N/3)^(1/2k)).)

So for k = 1 and N = 100 (i.e.) x^2+y^2+z^2 = 100 would result in a search space loosely bounded by

0 <= x <= 100^(1/2) = 10
0 <= y <= 100^(1/2) = 10
0 <= z <= 100^(1/2) = 10

a cube of 0 though 10 on each side or 11^3 = 1331 as a LOOSE bound. Not very hard. Higher powers result in even a smaller search space for a given N.