Harristotelian Logic

Working through the mathematics for surrogate factoring I found that for circles

a^2 + b^2 = r^2

with an integer radius greater than 4, there will be at least one integer solution for non-zero a and b, only about 1/3 of the time. As an example to understand the conjecture consider the first four cases:

3^2 + 4^2 = 5^2

6^2 + 8^2 = 10^2

5^2 + 12^2 = 13^2

and

8^2 + 15^2 = 17^2

the first four such solutions over the space from 5 to 17, which is 13 numbers, so you have 4/13 and an almost perfect 1/3 of the cases.

That follows from surrogate factoring as adding one more congruence to a congruence of squares gives the following:

x^2 = y^2 mod T

you can let k = 2x mod T so

2x = k mod T, multiply both sides by k, to get

2xk = k^2 mod T

and now add back to the original congruence of squares to get

x^2 + 2xk = y^2 + k^2 mod T

and then add k^2 to both sides and get

(x+k)^2 = y^2 + 2k^2 mod T

so you relate every factorization of one composite a factorization of some other composite, which I call the surrogate.

Actually trying to use this idea to factor you need explicit equations:

(x+k)^2 = y^2 + 2k^2 + omega*T

And you can factor your target T by instead factoring 2k^2 + omega*T which I call the surrogate.

Notice that for ANY difference of squares the surrogate factorization exists.

Now the obvious question since you have to pick k and omega is, how likely are you to factor a given target composite T with your arbitrary choice of k and omega?

You have three possibilities with the two main congruences:

x^2 = y^2 mod T and k^2 = 2xk mod T

1. Neither is satisfied.


2. They are both true for one prime factor or only some prime factors of T.


3. They are both satisfied.

If you consider ALL possible solutions modulo T for a given k, which sets the residue modulo T of x, you trace out a hyperboloid, as you're in a three dimensional space, and then, any non-zero omega you pick gives a circular slice out of that hyperboloid, and possibility 3. will give you integer solutions to that circle!!!

And with no reason to pick either possibility over the other, you get that last possibility about 1/3 of the time, which gives the circle conjecture mentioned above, which I just say is a conjecture to be fair.

It seems to me that the proof is obvious enough at this point, but for now I'll call it a conjecture.