CollegeAnon !LAbIRp9cT. ID: 2e5cb9 May 1, 2018, 8:53 a.m. No.5732   🗄️.is 🔗kun   >>5734

>>5730

I'm seeing all the f's look good when they are square. I'm thinking we could increment or decrement D (f function) until we get f as a square. Then (x+n) is the square root of f

CollegeAnon !LAbIRp9cT. ID: 2e5cb9 May 1, 2018, 10:41 a.m. No.5734   🗄️.is 🔗kun   >>5736

>>5732

I think that the gcd of (x+n,f) for the starting record and the gcd of (x+n,f) for the correct record are always related somehow. Usually the correct one is a factor of the starting one. Sometimes the correct is a multiple of the starting one. We could do recursive like this:

get e,n,d,x,a,b

get gcd(x+n,f)

factor that:

go through the factors (or multiples of the factors) as (x+n) to test if (x+n)^2 + c is square.

If nothing works, start iterating the original gcd.

???

Profit!