Tuesday, January 15, 2008
JSH: Diophantine equations underlying
I thought it worth mentioning that what I call surrogate factoring is just using two Diophantine equations:
x^2 =3D y^2 + pr_1
and
z^2 =3D y^2 + nT
where I connect the classical difference of squares with a target composite to factor T, with another difference of squares. My initial approaches actually had nT on top, and I'd derive the second difference of squares usually as
(x+k)^2 =3D y^2 + 2k^2 + nT
and factor 2k^2 + nT, in order to try and get y directly, and go back to the earlier equation and factor, which is why x and y are there, but my latest approach was to flip things a bit, and I found I could introduce a prime p—chosen—and that rather than focus on getting y exactly, I could get z mod p and y mod p, where z =3D x + ak, so the latest burst of postings occurred after a couple of changes in direction as I was having trouble explaining with theory how the equations behaved as a Diophantine system, going the other way.
Crucial in understanding the value in this approach is realizing that the second set of difference of squares has ALWAYS BEEN THERE but mathematicians have either not been aware of it, or they didn't know that it constrained the classical difference of squares, whether you acknowledge it or not.
That's the way mathematics works—the existence of the other factorizations can apply rules to the one that people concentrate on, whether they care if they are there or not.
Now that is the "pure math" aspect of what I do, as years ago I speculated about whether or not you could factor one number through factoring another which is the concept of surrogate factoring, and back then I had no idea that I was looking for a Diophantine system of two difference of squares. And I had no clue that I could find z mod p and y mod p of the classical system through this approach.
If you think it's a trivial thing to get answers for both equations with all integers, then try it.
If you do so, you will find that they follow rules:
z =3D (2=E1)^{-1}(1 + 2=E1^2)k mod p
k^2 =3D (=E1^2+1)^{-1}(nT) mod p
where =E1 is non-zero and found such that k exists, and k is coprime to nT.
And y =3D (1+2=E1^2)^{-1} z mod p, or y =3D -(1+2=E1^2)^{-1} z mod p.
So at the heart of all the discussion, where the issue has always been effectiveness of factoring there is this "pure math" result which is set of congruence relations controlling the Diophantine system of two differences of squares.
x^2 =3D y^2 + pr_1
and
z^2 =3D y^2 + nT
where I connect the classical difference of squares with a target composite to factor T, with another difference of squares. My initial approaches actually had nT on top, and I'd derive the second difference of squares usually as
(x+k)^2 =3D y^2 + 2k^2 + nT
and factor 2k^2 + nT, in order to try and get y directly, and go back to the earlier equation and factor, which is why x and y are there, but my latest approach was to flip things a bit, and I found I could introduce a prime p—chosen—and that rather than focus on getting y exactly, I could get z mod p and y mod p, where z =3D x + ak, so the latest burst of postings occurred after a couple of changes in direction as I was having trouble explaining with theory how the equations behaved as a Diophantine system, going the other way.
Crucial in understanding the value in this approach is realizing that the second set of difference of squares has ALWAYS BEEN THERE but mathematicians have either not been aware of it, or they didn't know that it constrained the classical difference of squares, whether you acknowledge it or not.
That's the way mathematics works—the existence of the other factorizations can apply rules to the one that people concentrate on, whether they care if they are there or not.
Now that is the "pure math" aspect of what I do, as years ago I speculated about whether or not you could factor one number through factoring another which is the concept of surrogate factoring, and back then I had no idea that I was looking for a Diophantine system of two difference of squares. And I had no clue that I could find z mod p and y mod p of the classical system through this approach.
If you think it's a trivial thing to get answers for both equations with all integers, then try it.
If you do so, you will find that they follow rules:
z =3D (2=E1)^{-1}(1 + 2=E1^2)k mod p
k^2 =3D (=E1^2+1)^{-1}(nT) mod p
where =E1 is non-zero and found such that k exists, and k is coprime to nT.
And y =3D (1+2=E1^2)^{-1} z mod p, or y =3D -(1+2=E1^2)^{-1} z mod p.
So at the heart of all the discussion, where the issue has always been effectiveness of factoring there is this "pure math" result which is set of congruence relations controlling the Diophantine system of two differences of squares.
# posted by James Harris @ 13:02 
