Yeah I know. 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. I've noticed that sometimes the correct (x+n) can be found by chopping off the last bit of e or f. Sometimes you take f*2 +/- 1 to get the correct answer. Still a lot to think about.