Wednesday, November 04, 2009
JSH: This factoring thing is scary
You people need to wake up. I'm telling you the underlying math is EASY. It's a technique to solve quadratic residues modulo N.
If you think it's not possible then use all of your mathematical training to explain the second example:
So now let's do, k=9. So q = 11 mod 35, and T = 22 mod 35, so I can try T = 57.
The trivial factorization didn't work here, so I'll just jump to, f_1 = 19, and f_2 = 3, so:
k = 3^{-1}(19 + 3) mod 35 = 12(22) mod 35 = 264 mod 35 = 19 mod 35.
19^2 = 11 mod 35
so it worked! (It's so weird though watching it. Even though I know the underlying mathematics it seems like magic.)
And that is a factoring example, as I know k=9 is a solution, so I have
19^2 = 9^2 mod 35
so (19-9)(19+9) = 0 mod 35, so (10)(28) = 0 mod 35, and you pull 5 and 7 as factors with a gcd.
THAT is how you use a method for solving quadratic residues modulo N: you find one quadratic residue and then go looking for another.
Factoring problem solved.
Happy one year birthday to the solution as it's a year old about now.
If you think it's not possible then use all of your mathematical training to explain the second example:
So now let's do, k=9. So q = 11 mod 35, and T = 22 mod 35, so I can try T = 57.
The trivial factorization didn't work here, so I'll just jump to, f_1 = 19, and f_2 = 3, so:
k = 3^{-1}(19 + 3) mod 35 = 12(22) mod 35 = 264 mod 35 = 19 mod 35.
19^2 = 11 mod 35
so it worked! (It's so weird though watching it. Even though I know the underlying mathematics it seems like magic.)
And that is a factoring example, as I know k=9 is a solution, so I have
19^2 = 9^2 mod 35
so (19-9)(19+9) = 0 mod 35, so (10)(28) = 0 mod 35, and you pull 5 and 7 as factors with a gcd.
THAT is how you use a method for solving quadratic residues modulo N: you find one quadratic residue and then go looking for another.
Factoring problem solved.
Happy one year birthday to the solution as it's a year old about now.