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Frederick Williams

Joined: 19 Nov 2005
Posts: 97 Posted: Sat Jul 08, 2006 1:35 am    Post subject: Re: New factoring idea jstevh@msn.com wrote:
 Quote: Frederick Williams wrote: jstevh@msn.com wrote: ... And just like that a bit of amateur mathematical research showing what I hope is a new way to factor. Assuming that it is new, how about showing that it's _useful_? Try factoring some big numbers with it. Well, I'll admit that having had so many failed ideas, and with it being such a letdown to figure out something doesn't work well, I like to fully enjoy the part where I focus on it being a new idea, and have hope that it could be important.

You mean it being new is more important than it being right? Look
here's something new:

7654854 + 89723648763 = 796596876486.

(Quick proof that it's new: it doesn't appear in Wikipedia.)

It's not right, but according to you it has some value because it's new,
what value exactly?

 Quote: ... look back over that factoring idea and ask yourself, can you be sure that mathematicians ignoring it means it does not work well?

I think the onus is on you to prove that it works well.

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Remove "antispam" and ".invalid" for e-mail address. jstevh@msn.com
science forum Guru

Joined: 21 Jan 2006
Posts: 951 Posted: Sat Jul 08, 2006 12:10 am    Post subject: Re: New factoring idea Frederick Williams wrote:
 Quote: jstevh@msn.com wrote: ... And just like that a bit of amateur mathematical research showing what I hope is a new way to factor. Assuming that it is new, how about showing that it's _useful_? Try factoring some big numbers with it.

Well, I'll admit that having had so many failed ideas, and with it
being such a letdown to figure out something doesn't work well, I like
to fully enjoy the part where I focus on it being a new idea, and have
hope that it could be important.

And besides, that's a check of the mathematical community.

If this idea IS an excellent one, what would it mean for it to be
ignored?

Remember this is just one more bit of research for me, like I found my
own prime counting function, and in frustration one day after arguing
with math people who kept fighting its uniqueness and having noticed
that the Wikipedia didn't have a prime counting function article (it
had a re-direct to a prime theorem page), I wrote the first prime
counting function for the Wikipedia.

After some iterations and a bit of help from a couple of other people,
I had the page you can see in the history which shows you some of my
OTHER number theory research:

http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=9142249

Now that is a one-line definition for a prime counting function that
not only counts prime numbers it finds prime numbers as it recurses,
unlike any other known.

Mathematicians refuse to fully acknowledge my research and on Usenet,
math people routinely lie about it, and call me names.

I suggest to all of you that it would not change if you had the results

It's a community problem.

Here with the factoring problem, you can see how bad it is.

These math professors are playing political games with mathematics, and
I say, playing with the future of humanity.

Look over my prime counting function in the Wikipedia article I wrote,
and then look back over that factoring idea and ask yourself, can you
be sure that mathematicians ignoring it means it does not work well?

And if you can, try to imagine how I feel, how devastating it is to
make discoveries and face such repudiation and insults when you
accomplished something.

James Harris a.khanm@gmail.com
science forum beginner

Joined: 02 Jul 2006
Posts: 14 Posted: Fri Jul 07, 2006 8:02 am    Post subject: Re: New factoring idea Please apply your method to non-trivial cases and post the results
here. I believe that in doing so you will realize the problems with
this method.

jstevh@msn.com wrote:
 Quote: jstevh@msn.com wrote: 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 so of course I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!! But how do you find 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!!! 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. And just like that a bit of amateur mathematical research showing what I hope is a new way to factor. In a world where you'd think some people care about mathematics. James Harris Patrick Hamlyn
science forum beginner

Joined: 03 May 2005
Posts: 45 Posted: Fri Jul 07, 2006 3:04 am    Post subject: Re: New factoring idea Frederick Williams <Frederick.Williams1@antispamtesco.net.invalid> wrote:

 Quote: jstevh@msn.com wrote: ... And just like that a bit of amateur mathematical research showing what I hope is a new way to factor. Assuming that it is new, how about showing that it's _useful_? Try factoring some big numbers with it.

Jeeeezzz... don't you remember what happened last time somebody asked that?

Let's not go through that all over again... Frederick Williams

Joined: 19 Nov 2005
Posts: 97 Posted: Fri Jul 07, 2006 2:59 am    Post subject: Re: New factoring idea jstevh@msn.com wrote:
 Quote: ... And just like that a bit of amateur mathematical research showing what I hope is a new way to factor.

Assuming that it is new, how about showing that it's _useful_? Try
factoring some big numbers with it.

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Remove "antispam" and ".invalid" for e-mail address. jstevh@msn.com
science forum Guru

Joined: 21 Jan 2006
Posts: 951 Posted: Thu Jul 06, 2006 5:50 am    Post subject: Re: New factoring idea jstevh@msn.com wrote:
 Quote: 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 so of course I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!! But how do you find 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!!!

Now you can get k.

 Quote: 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. And just like that a bit of amateur mathematical research showing what I hope is a new way to factor.

In a world where you'd think some people care about mathematics.

James Harris jstevh@msn.com
science forum Guru

Joined: 21 Jan 2006
Posts: 951 Posted: Thu Jul 06, 2006 12:00 am    Post subject: Re: New factoring idea Ulrich Diez wrote:
 Quote: James Harris wrote: ... x^2 - a^2 = S + T and x^2 - b^2 = S - k*T subtract the second from the first to get b^2 - a^2 = (k+1)*T which is ... a factorization of (k+1)*T: ... Let's number the equations: I) x^2 - a^2 = S + T II) x^2 - b^2 = S - k*T I)-II) -> III) b^2 - a^2 = (k+1)*T ... But how do you find 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 ... If I got it right, you want an integer-factorization of T. (a and x will only be integers as long as f_1 and f_2 are of same parity.)

I take it you mean if f_1 and f_2 are both odd or both even, which is
correct.

 Quote: So you chose S more or less arbitrarily in a way so that the right side (S+T) of equation I can easily be written as product of two integer-factors of same parity/as difference of two squares for obtaining integer values for a and x. This means that while dealing with equation I, you choose/ calculate integer-values for a, x, S and T. After having chosen/calculated these values, you need to solve II for integer b and k. Why do you think that such integer b and k exist at all for any more or less arbitrarily chosen S? If a,b,x,S,T are not all integer - what makes you think that you can easily obtain an integer-factorization of T from your equation-system? Or looking at it from the other side: How to find S so that - both S+T can easily be factorized - and there are all-integer x,a and b that will solve your equation-system ?

I step through how in the original post.

You end up with a congruence relationship, solve it, and you're done.

It's weirdly simple.

The gist of it, is that if you have g_1*g_2 = S+T, and f_1*f_2 = S -
k*T, you find values for k such that

g_1 + g_2 = f_1 + f_2

and doing so is just a matter of solving a congruence equation, where
quadratic residues step in, as I show in my original post.

Some remarkable mathematics that you have to keep staring at to believe
it, as it just looks too simple, even to me.

How could it be so simple, and yet not be already known?

James Harris Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593 Posted: Wed Jul 05, 2006 10:42 pm    Post subject: Re: New factoring idea jstevh@msn.com wrote:
 Quote: Was doodling, playing around with some simple equations [ Derivation snipped ] Really simple so of course I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!! [...]

In all seriousness, James, the dead ends of research are NOT published
(in general). So just because this approach isn't mentioned, doesn't
mean that no one thought of it.

For instance, the Four Color Theorem, which states that in any map, you
can color the regions with four colors such that no two adjacent
regions get the same color. It is not too difficult to show that you
cannot have five countries, all adjacent to each other. A COMMON
mistake, which is probably made every 6 months or so, is that this is
the same as the 4CT. However, it is not; and you will not find this
fact at Wikipedia, or even at MathWorld.

In fact, someone may have been able to prove that that particular
approach cannot work. But since it's something that doesn't work, it
wouldn't have been publishable, and no mention would have been made in
Wikipedia.

--- Christopher Heckman Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593 Posted: Wed Jul 05, 2006 10:34 pm    Post subject: Re: New factoring idea Frederick Williams wrote:
 Quote: jstevh@msn.com wrote: ... I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!! Meaning what? That because it isn't mentioned in Wikipedia, it must be new?

It must be, according to JSH's logic.

--- Christopher Heckman Ulrich Diez
science forum beginner

Joined: 01 Nov 2005
Posts: 9 Posted: Wed Jul 05, 2006 4:49 pm    Post subject: Re: New factoring idea James Harris wrote:

....
 Quote: x^2 - a^2 = S + T and x^2 - b^2 = S - k*T subtract the second from the first to get b^2 - a^2 = (k+1)*T which is ... a factorization of (k+1)*T: ....

Let's number the equations:

I) x^2 - a^2 = S + T
II) x^2 - b^2 = S - k*T
I)-II) -> III) b^2 - a^2 = (k+1)*T

....
 Quote: But how do you find 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 ....

If I got it right, you want an integer-factorization
of T.
(a and x will only be integers as long as f_1 and f_2
are of same parity.)
So you chose S more or less arbitrarily in a way so that
the right side (S+T) of equation I can easily be written
as product of two integer-factors of same parity/as difference
of two squares for obtaining integer values for a and x.
This means that while dealing with equation I, you choose/
calculate integer-values for a, x, S and T.

After having chosen/calculated these values, you need to
solve II for integer b and k.
Why do you think that such integer b and k exist at
all for any more or less arbitrarily chosen S?
If a,b,x,S,T are not all integer - what makes you think
that you can easily obtain an integer-factorization of T
Or looking at it from the other side:
How to find S so that
- both S+T can easily be factorized
- and there are all-integer x,a and b that will solve

Ulrich fishfry
science forum Guru Wannabe

Joined: 29 Apr 2005
Posts: 299 Posted: Wed Jul 05, 2006 2:07 am    Post subject: Re: New factoring idea jstevh@msn.com wrote:

 Quote: Really simple so of course I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!!

Right. If it's not on Wiki, nobody ever thought of it before.

James, why not spend the time learning some math? Frederick Williams

Joined: 19 Nov 2005
Posts: 97 Posted: Wed Jul 05, 2006 1:52 am    Post subject: Re: New factoring idea jstevh@msn.com wrote:
 Quote: ... I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!!

Meaning what? That because it isn't mentioned in Wikipedia, it must be
new?

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Remove "antispam" and ".invalid" for e-mail address. jstevh@msn.com
science forum Guru

Joined: 21 Jan 2006
Posts: 951 Posted: Wed Jul 05, 2006 12:34 am    Post subject: New factoring idea 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 so of course I'm thinking it's not new, but looking over at the Wikipedia: http://en.wikipedia.org/wiki/Factoring It's new!!! But how do you find 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. 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. And just like that a bit of amateur mathematical research showing what I hope is a new way to factor. James Harris  Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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