Harristotelian Logic

Thought I'd make a cleaner post without errors showing this simple factoring idea that I just figured out a couple of days ago.

It IS surrogate factorization which can be seen from the start, as S+T is the first factorization equation shown, and the math is straightforward from there:

x^2 - a^2 = S + T

and

x^2 - b^2 = S - k*T

I could subtract the second from the first to get

b^2 - a^2 = (k+1)*T

which is, of course, a factorization of (k+1)*T:

(b - a)*(b+a) = (k+1)*T

with integers for S and T, where T is the target composite to factor, so you have to pick this other integer S, and factor S+T.

Really simple.

But how do you find all the variables?

Well, if you pick S, and have a T you want to factor, then using

f_1*f_2 = S+T

it must be true that

a = (f_1 - f_2)/2

And

x=(f_1 + f_2)/2

so, you need the sum of factors of (S-k*T)/4 to equal the sum of the factors of (S+T)/4, so I introduce j, where

S - k*T = (f_1 + f_2 - j)*j

and now you solve for k, to get

k = (S - (f_1 + f_2 - j)*j)/T

so you also have

S - (f_1 + f_2 - j)*j = 0 mod T

so

j^2 - (f_1 + f_2)*j + S = 0 mod T

and completing the square gives

j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T

so

(2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T

so you have the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to find j, which is kind of neat, while it's also set what the quadratic residue is, so there's no search involved.

The main residue is a trivial result that gives k=-1, but you have an infinity of others found by adding or subtracting it from multiples of T.

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It was pointed out to me that these are also trivial, so I figured out a way around that by turning the problem around a bit:

One approach is to find some quadratic residue r, where

(f_1 + f_2)^2 - 4*S = r + n*T

where n is a natural number, as then solving for f_1 gives

f_11 = (sqrt(4*S + r + n*T) +/- sqrt(r + (n-1)*T))/2

so you can arbitrarily pick some integer w, square it, and get the quadratic residue modulo T, which is then your r, so now you have

w^2 = r + (n-1)*T

so you can easily solve for n, and then you pick S so that the second square root is an integer.

So now you have

2*j - (f_1 + f_2) = w

is a solution.

Neat!!! I like solving problems!!!

Tell me more!

Now you can get k.

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And then you can find b, from

b^2 = x^2 - S + kT

and you have the factorization:

(b-a)*(b+a) = (k-1)*T.

It is possible to generalize further using

j = z/y

and then the congruence equation becomes

(2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T.

If you're skeptical you may consider the question of finding k when you already have the factorization of T.

I've been thinking about this and looking at it, and it looks like factoring is easy—if you know how to do it—where the lateral thinking step needed was to start with equations factoring your target added to something else.

Weird. I wonder why Gauss didn't think of this?

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It is a bit harder than I first realized though.

Hey, do you people think that maybe sitting quiet may be like being a frog sitting still in a pot water that is on a fire?

Did it ever occur to any of you that I've had over three years to get rather pissed off, since you people blocked my other research, like my proof of Fermat's Last Theorem?

I WILL empty entire math departments, maybe starting with Princeton.