### Sunday, June 11, 2006

## SF: Square roots were the key?

For years I've been looking for a way to factor one number using the factorization of some different number, which I call the surrogate for a method I call surrogate factoring or SF.

Trouble is, every set of equations I played with would either factor pathetically, as in with a very low success rate, or I'd find that I'd just go in a circle needing the factorization of my target.

As a very simple example that isn't exactly what I used consider something like

T = (3x - 2y)(x+2y)

so if you solve it out for a factorization you have

3x - 2y = f_1, and x + 2y = f_2

well, of course, if you solve the thing, some of the ways I was doing, where'd I'd do completions of the square, you just get what you'd have if you solve out the two linear equations.

Well I pondered that for a while, and wonder, what if I give the algebra an ambiguous expression that can't have a single solution?

So after a bit I came up with

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

where S is the surrogate, so that the math does NOT see the target T at all, for this first step, so it can't make everything rely on the factorization of T, but the expression is ambiguous because of the square roots.

Yet, of course, it's trivial to find squares that will fit in for some S that you pick, like S=15.

So what do you do with the thing?

You multiply out, and get rid of the radicals by squaring to get handle terms with sqrt(xy), and then you complete the square twice.

Someone posted a result of that on sci.math where that person used math software which is a good idea as it's a mess.

Next you focus on the term that now doesn't have x or y, and based on what was posted on sci.math, it will have S^2 as a factor, and a lot of complicated stuff with k_1, k_2, k_3, and k_4, and you take that complicated stuff and set it to T, your target composite to factor.

If that person's result was correct, then if you can use math software, I'd recommend it.

Next part is where I don't know as you have k's raised to the 4, but you have 4 degrees of freedom, meaning you can set at least two of the k's to just about anything, other than 0, and then figure out what value for the next to last, will give you an integer for the last one.

If you can manage that—that's the hard part, where if it can't be done practically, then this is yet another pure math curiousity—then you just go back to

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

and you can even use, like, S=15, and then you solve for x and y, like

k_1*sqrt(x) + k_2*sqrt(y) = 5

k_3*sqrt(x) + k_4*sqrt(y) = 3

and with x and y, you can I think just switch signs there, with the factors of your target popping out, or you can go back to that messy expression and plug in your values for x and y to get back a factorization of your target.

The target gets incidentally factored.

If it can be made to work, it looks like it'll take a lot of expertise far outside what I usually do, as I focus on rather simple algebra most of the time.

The hardest thing being getting integer k's out of that monster expression where the highest exponent is 4.

Though I guess even that may require factoring T, so the algebra could still force you to know the factorization of T anyway, which I suspect is what happens, as that is what usually shoots down my surrogate factoring ideas.

They kind of just go in a big freaking circle what I call a BFC, where you need the factorization of T anyway.

Trouble is, every set of equations I played with would either factor pathetically, as in with a very low success rate, or I'd find that I'd just go in a circle needing the factorization of my target.

As a very simple example that isn't exactly what I used consider something like

T = (3x - 2y)(x+2y)

so if you solve it out for a factorization you have

3x - 2y = f_1, and x + 2y = f_2

well, of course, if you solve the thing, some of the ways I was doing, where'd I'd do completions of the square, you just get what you'd have if you solve out the two linear equations.

Well I pondered that for a while, and wonder, what if I give the algebra an ambiguous expression that can't have a single solution?

So after a bit I came up with

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

where S is the surrogate, so that the math does NOT see the target T at all, for this first step, so it can't make everything rely on the factorization of T, but the expression is ambiguous because of the square roots.

Yet, of course, it's trivial to find squares that will fit in for some S that you pick, like S=15.

So what do you do with the thing?

You multiply out, and get rid of the radicals by squaring to get handle terms with sqrt(xy), and then you complete the square twice.

Someone posted a result of that on sci.math where that person used math software which is a good idea as it's a mess.

Next you focus on the term that now doesn't have x or y, and based on what was posted on sci.math, it will have S^2 as a factor, and a lot of complicated stuff with k_1, k_2, k_3, and k_4, and you take that complicated stuff and set it to T, your target composite to factor.

If that person's result was correct, then if you can use math software, I'd recommend it.

Next part is where I don't know as you have k's raised to the 4, but you have 4 degrees of freedom, meaning you can set at least two of the k's to just about anything, other than 0, and then figure out what value for the next to last, will give you an integer for the last one.

If you can manage that—that's the hard part, where if it can't be done practically, then this is yet another pure math curiousity—then you just go back to

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

and you can even use, like, S=15, and then you solve for x and y, like

k_1*sqrt(x) + k_2*sqrt(y) = 5

k_3*sqrt(x) + k_4*sqrt(y) = 3

and with x and y, you can I think just switch signs there, with the factors of your target popping out, or you can go back to that messy expression and plug in your values for x and y to get back a factorization of your target.

The target gets incidentally factored.

If it can be made to work, it looks like it'll take a lot of expertise far outside what I usually do, as I focus on rather simple algebra most of the time.

The hardest thing being getting integer k's out of that monster expression where the highest exponent is 4.

Though I guess even that may require factoring T, so the algebra could still force you to know the factorization of T anyway, which I suspect is what happens, as that is what usually shoots down my surrogate factoring ideas.

They kind of just go in a big freaking circle what I call a BFC, where you need the factorization of T anyway.