View Single Post
Old 2021-11-16, 14:18   #2
R. Gerbicz
R. Gerbicz's Avatar
"Robert Gerbicz"
Oct 2005

30738 Posts

For any p the e values is not lower bounded:
set say x=1, then x^2*(x^3+y^3)=y^3+1 goes to -inf for y->-inf.

For your 2nd question: about solving the equation for a given n=p+e:
first method: factorize n, then try each squared divisor as for d=x^2 you can solve it: y=(n/d-x^3)^(1/3) ofcourse check if y is an integer or not (you had two choices for x: x=+-sqrt(d)).
2nd method: economical solution, find only all p<10^9 that is a prime divisor of n (and then the exact primepower p^e divisor), after this again try the squared divisors, the idea of this approach is that for random(!) n values it is unlikely that there is even a single p value for that p^2|n (you have less than 1e-9 probability for this).
R. Gerbicz is offline   Reply With Quote