Saturday, June 10, 2006
JSH: Pure math, factoring idea
Well if this latest idea of mine actually can be made to work, it will be so damn ironic that it will be hard to comprehend that irony.
Over the years I've gotten into stupid arguments with math people on this forum about the inherent ambiguity of the sqrt() as it HAS to give two answers, no matter what people define.
Well, looking at
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
I used that reality in a very, very clever way, if it works!!!
If you multiply everything out, eliminate the radical by squaring, and then complete the square twice, you should get this rather complicated expression, which is what is factored by solutions for x and y.
Being a clever person you pick S as your surrogate, so you know squares for x and y that will work!!!
But then you get a bonus! Because there are TWO solutions to the square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless there's some block as I haven't looked at the complicated expression) you may end up factoring a target composite.
If this idea works, you might be able to factor an RSA sized number using S=15.
And that would also be rather ironic.
I guess I really should have math software that can do that calculation. I'm not even going to try it just yet, hoping someone else will, as I'm so likely to make errors doing it by hand.
It is a bit of a horrendous calculation to do by hand.
[A reply to someone who said that the calculation is trivial.]
Still I make dumb mistakes with such calculations.
I take it then you found that k_1, k_2, k_3 and k_4 are related to the factorization of your target?
That's what I expect to be true.
It is kind of neat, pure math kind of thing though. A way to express two factorizations using just one.
I guess none of you want to play along any more which is why no one posted the answer.
And you really shouldn't do such a calculation by hand when math software is so good at it, and—perfect.
Over the years I've gotten into stupid arguments with math people on this forum about the inherent ambiguity of the sqrt() as it HAS to give two answers, no matter what people define.
Well, looking at
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
I used that reality in a very, very clever way, if it works!!!
If you multiply everything out, eliminate the radical by squaring, and then complete the square twice, you should get this rather complicated expression, which is what is factored by solutions for x and y.
Being a clever person you pick S as your surrogate, so you know squares for x and y that will work!!!
But then you get a bonus! Because there are TWO solutions to the square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless there's some block as I haven't looked at the complicated expression) you may end up factoring a target composite.
If this idea works, you might be able to factor an RSA sized number using S=15.
And that would also be rather ironic.
I guess I really should have math software that can do that calculation. I'm not even going to try it just yet, hoping someone else will, as I'm so likely to make errors doing it by hand.
It is a bit of a horrendous calculation to do by hand.
[A reply to someone who said that the calculation is trivial.]
Still I make dumb mistakes with such calculations.
I take it then you found that k_1, k_2, k_3 and k_4 are related to the factorization of your target?
That's what I expect to be true.
It is kind of neat, pure math kind of thing though. A way to express two factorizations using just one.
I guess none of you want to play along any more which is why no one posted the answer.
And you really shouldn't do such a calculation by hand when math software is so good at it, and—perfect.
# posted by James Harris @ 22:58 
