>>5470

I'm thinking that we need to generate a pythagorean triple with a^2+b^2=c^2 where a or b is our original number squared. Then we can use this to get:

c^2 - b^2 = a^2. We can then set our D and X vaues as c and b respectively, then we get:

D^2 - X^2 = a^2

Then we can use this to get the correct little d value = 2D-c and correct little x = 2X + c.

>>5474

http://fermatslibrary.com/p/56b57e1c

This is supposedly how Fermat generated the triples.

I'm trying to look at D,X,C where D^2-X^2=C^2. I've done it for a bunch of primes (mod 60) [60 = 345]. and I've generalized it into a few cases:

(D,X,C) MOD 60:

i) (1,0,c) c in {1,11,19,29,31,41,49,59}

ii) (25,24,c) c in {7,13,17,23,37,43,47,53}

iii) (25,36,c) c in {7,13,17,23,37,43,47,53}

iv)(49,0,c) c in {1,11,19,29,31,41,49,59}

So if c is in {1,11,19,29,31,41,49,59} we know that X is divisible by 60. If C is in {7,13,17,23,37,43,47,53} then we know that D-25 is divisible by 60. If we can get this D and X then we can use algebra to solve for d^2-x^2=c. We get this triple somehow, then we have a big D. Then we know that our D value can be split up into a sum of two squares because D = a^2 + b^2 and then get the squares by Fermat.