The new strong (extended) factoring method using the 4-square representation of an integer.

[mahbel]

Originally Posted by science_man_88

you said 4 square representations to another not a mixed representation. to get back squares you need a sum of squares that sum to a square the simplest form is a pythagorean triple.

so 1^2+3^2+4^2+5^2 =1^2+5^2+5^2+0^2 because 3^2+4^2=5^2.

my post showed that the expansion of squares (a,b,c,d) into sum of squares helps in response to your post in which you said it doesn't. And I gave an example which clearly shows that the expansion helps. there are two different things.

1-that one 4-sq rep can be transformed into another 4-sq rep. And my previous comment was not about this.
2-that the expansion of elements of a 4-sq rep (a,b,c,d) can themselves be expanded into sums of 4-sq then those expansions can be used to make combinations that lead to factors.

The example I provided (5,5,5,4) clearly showed that the expansion helps greatly because it allowed us to find factors more often than in just considering the original (5,5,5,4) to find factors.

We don't really need a sum of square that produce Pythagorean triplets. Yes, they may appear in any expansion but more often than not, they don't.

[CRGreathouse]

I still think this is an artifact of working with small numbers. It's not clear to me that you could produce a lot of 4-square representations for a moderate-size number given one such representation.

[Dr Sardonicus]

Originally Posted by mahbel

The example I provided (5,5,5,4) clearly showed that the expansion helps greatly because it allowed us to find factors more often than in just considering the original (5,5,5,4) to find factors.

With a number of size less than 100 whose factors are less than 16, and a set of 16 numbers from which you can form the sums of any of its 65536 subsets, I'd say you have a decent chance of hitting a multiple of either of its factors.

Just to review, in the very first post to this thread here you said:

We start with the well known representation of an integer N as a sum of 4 squares. It will be shown that it is possible to extract the factors of N themselves if we used the representation of N or 2N as a sum of 4 squares in a straightforward and efficient way. The following example will show how the method is implemented.

We consider the representation of 2N as a sum of 4 squares (written as (x,y,z,t) without the exponent 2) and we will show below that we can in fact extract factors of 2N. We will see that we can easily find pairs or triplets of squares, which when added together, will make up numbers M(i) smaller than N ( since we gain nothing by adding the 4 squares ) that may share a factor with 2N.

Now, you're saying

It turned out that one 4-sq representation can be transformed into another simply by expanding the squares in the first one and re-arranging them to create new 4-sq representations.

So, you're changing your "method" yet again. And, I ask -- "new 4-sq representations" of what, exactly? Not of the original number, it would seem.

And, when challenged with a number of any size to test your new "method" on, you again refuse the challenge, whining that

I cannot do the calculations by hand.

This is ridiculous. If you can't work with numbers of any size, and give no theoretical basis for your claims, you have no right to make them.

Let's see here. You've gone from 4 squares to, now, 16 squares. And, of course, the overhead of computing four additional 4-square representations, for each representation of the original number. If your original method was slow, this is slow to the fifth power! Let's see. How did you describe your method originally? Straightforward? Efficient?

[mahbel]

Originally Posted by CRGreathouse

I still think this is an artifact of working with small numbers. It's not clear to me that you could produce a lot of 4-square representations for a moderate-size number given one such representation.

You may be right. It could be just an artifact. But as long as we don't have a proof, we can't rule it out.

[CRGreathouse]

Originally Posted by mahbel

You may be right. It could be just an artifact. But as long as we don't have a proof, we can't rule it out.

Right, just like we can't rule out that the good doctor's scattergun won't instantly factor most large semiprimes just beyond the range we've tested.

[Dr Sardonicus]

Originally Posted by mahbel

You may be right. It could be just an artifact. But as long as we don't have a proof, we can't rule it out.

From The PrimeNumbers' Crackpot index

10 points for expecting others to disprove your result(s) rather than providing the proof yourself.
30 points for confusing examples and/or heuristics with mathematical proof.
50 points for failing to respond to appropriate corrections, questions and challenges.