
#1
Feb2013, 06:33 PM

P: 1

Hello,
I'm currently learning all about factorization techniques for a college project. I know that most of the advanced techniques are based around a congruence of squares. As part of our research we've been asked to think up an original factoring method. It doesn't matter how slow or fast it is as long as its original, and we get bonus points for this if we can prove it's original. I have thought of a method, although I don't think it is original, but I'm not sure, so I've come on here to check. I'll give the method with some examples which I've been working through in excel. Tbh I've been working through many examples, and I've hit the number limit with excel which is above 10 digits or something. Anyway, here goes: Example (1): Factorize the number 77. Method: 77 = 8^2 + 13 138 = 5, 5^2 + 4.13 = 77 Now we have that 77 = 8^2 + 1.13 ...(1) and 77 = 5^2 + 4.13 ...(2) Now multiply (1) by 2^2, giving 308 = 16^2 + 4.13 ...(3) Then subtract (2) from (3) giving 231 = 16^2  5^2 Then factorise 77 as gcd(77, 165)*gcd(77,16+5) i.e. 77 = 11 * 7 Example (2): Factorize the number 8791. Unlike with 77, which worked on the first attempt, this only works at the third attempt. Each attempt multiply the number in question by 1,2,3,4,5... So attempts 1 and 2 didn't result in a square being generated, but attempt 3 does: 3*8791 = 26373 = 162^2 + 129 Now do Abs(129  162) = 33 33^2 + 196.129 = 26373 So we have that: 162^2 + 1.129 = 26373 ...(1) 33^2 + 196.129 = 26373 ...(2) Now multiply 1 by 14^2 (=196) to give: 2268^2 + 196.129 = 196.26373 ...(3) And: 33^2 + 196.129 = 26373 Subtracting (2) from (3) gives: 2268^2 33^2 = 195.26373 Thus 8791 = gcd(226833, 8791)*gcd(2268+33,8791) = 149 * 59 Example (3): Factorize 8414786257. A square was found in excel at the 1496th attempt. 1496*8414786257 = 3548030^2 + 3359572 = 188458^2 + 3736489.3359572 i.e. 1496*8414786257 = 3548030^2 + 3359572 = 188458^2 + (1933^2).3359572 Thus gcd(1933.3548030188458, 8414786257) = 104297 and gcd(1933.3548030+188458, 8414786257) = 80681 Q.E.D. I hope someone will be able to give me some feedback on this. As I say, I don't think this method can possibly be fully original, but I may get partial credit for any original element? Thanks for your help in this. :D I'm running into a bit of a problem with excel, because of floating point integers and the limits with representation. I'd like to test out this method on larger integers, maybe using C. I'm a bit rusty on it at the moment, so if anyone could implement it fairly quickly and let me know how the results shape up I'd be grateful. I've reached the limit with what can be done with excel. This is the largest number I've been able to factorize thus far: 30,436,307,070,163. It's only 14 digits long. I want to test this method with semiprime numbers up to 20 or 30 decimal digits at least. 30,436,307,070,163 was factorized as follows. At the 10,560th attempt a square was found: 10,560*30,436,307,070,163 = 566,928,040^2 + 122,679,680 = 444,248,360^2 + (31,799^2)*122,679,680 Thus gcd(31,799*566,928,040444,248,360; 30,436,307,070,163) = 8,589,337. The other factor is gcd(31,799*566,928,040+444,248,360; 30,436,307,070,163) = 3,543,499 From previous posts on this site, I've seen somewhere that a method must be tested for numbers greater than 2^64 to see if it is truly effective, and this falls well short of that, which is about 20 decimal digits, although this last factorization is much better than trial division, since sqrt(30,436,307,070,163) = 5,516,911 and there are about 382,000 primes below this number, hence 10,560 attempts is not bad at this stage. However I think this will be insufficient as the numbers grow larger, but I have no idea as I lack the ability to test this out on C at the moment. If anyone could quickly check this out by seeing how it performs with two 15 digit primes, spaced not too close together, I would be very grateful to them. 



#2
Feb2113, 07:15 AM

P: 84

you write:
5^2 + 4.13 =77 that's wrong. 5^2 + 4.13 =25+4.13 =29.13 same thing with your equations 1 and 2. How can 77 be equal to 8^2 + 1.13 ? and what exactly do you mean by ... in equation 1 and 2? 



#3
Feb2113, 08:13 AM

P: 891





#4
Feb2113, 08:18 AM

P: 891

Factoring Method
I havn't seen this method of factoring before. However, proving it original may be a problem. You may try the wolfram site to work math problems with large numbers.




#5
Feb2113, 08:26 AM

P: 234

It seems to me that it is a modified form of Fermat factoring. A^2  B^2 Is the next to the last step.




#6
Feb2113, 08:48 AM

Emeritus
Sci Advisor
PF Gold
P: 16,101





#8
Feb2113, 10:07 PM

P: 84

Here's a list of factorization methods
http://en.wikipedia.org/wiki/Integer_factorization you may want to compare your method to the ones above. 


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