I want to define the frequency, p(n), to be the amount of times (in an integral domain) where a function happens to cube. For instance, f(x) = x^2 happens to be a cube when (x=1, x= 8, x = 27, x = 64), the frequency is thus 4/64 = 1/16 and happens to follow the relation [(n^(0.333333)]/(n) = p(n), where n is the number of integers in the domain, and p(n) is the frequency. The frequency function doesnt seem to change with an increasing n. however a small tweak to the function can actually have profound effects on the p(n) value and the adding of an extra term can make it much more complex. For instance, f(x) = x^2 + x^3 = y^3 (where x and y are integers) and we can actually create an upper bound for the frequency, not a good one, but better than 1. If x^2 is a cube, then it goes against fermat's last theorem, and we can eliminate all these possibilities as such, (1- ((n^0.3333333)/n) > p(n) a lower bound is not so obvious but can also be constructed i would guess, possibly because the function has little to share intersections with the fermat's last theorem eqn. of course, now it becomes about probability. any ideas?