Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: Factoring idea
Posted:
Jul 6, 2006 1:53 AM
|
|
jstevh@msn.com wrote:
> After years of effort with lots of failures I noticed something
> remarkably simple, and this time the math is all worked out, and there
> are few places for errors to hide.
>
> Was doodling, playing around with some simple equations and noticed
> that with
>
> 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 T.
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.
>
> 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.
>
Unfortunately, my opinion is that most of you could not become real
mathematicians, and there are too many people already who need to be
cleared out who have pretended for too long.
The speed here is slow to make a point.
And part of that point is that while sure, a lot of you might be
material for math teachers or math scholars, few if any of you can
actually be a mathematician.
It takes more than some degree at some school.
You have to be someone at least a little like me, or the people who
came before me on whose shoulders I stand and in whose memory I will
act as needed.
And for most of you that means you will never be mathematicians.
James Harris
|
|
|
|
