>>5740
Good points, CA! I'm examining them closely to make sure I understand your ideas. Here we go.
>We can solve (x+n) from f if f is a square number.
>If f is square, then (x+n) is the root of f.
>this leads me to believe that in order to factor c we need to factor f (recursive solution). This is because for our first step if we had a case where e = 0 then we'd have a solution. For this if we (for our f) have e=0, then we have a solution.
>I think we need to do this because the correct (x+n) often seems to be a multiple of a factor of f.
(often a multiple of the gcd of the original (x+n) and f).
>If f is prime, not sure what to do.
>Still a lot to think about.
Now we're talking. (e,1) is supposed to be our big clue. We still haven't figured that out. After 6-7 months. How does (e,1) play into factoring and given c or x+n value?