Comment on An exponent one-fifth algorithm for deterministic integer factorisationparentComments−archgoon5yThe abstract is correct. A positive integer N is log(N) bits long. If X = log(N), than a algorithm that runs in N^(1/5) will run in 2^(X/5) time.
Comments
The abstract is correct. A positive integer N is log(N) bits long. If X = log(N), than a algorithm that runs in N^(1/5) will run in 2^(X/5) time.