20170624, 00:41  #133  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010101100100_{2} Posts 
Quote:
http://www.mathcelebrity.com/foursqu...quare+Notation and factor it. Just do it yourself. Simply because you think that "people think that you are cheating" doesn't mean that you are not cheating. 

20170624, 00:57  #134  
Aug 2006
175B_{16} Posts 
Quote:
I've chosen two random semiprimes such that one is a sum of two squares and the other is not. They both have 120 decimal digits, which is designed to be large enough that they are inconvenient to factor. Can you tell, via this method, which is the sum of two squares? 212516552112132255144219181942145041853904419406206147277341101918011712262187398404745415408630814953685178049763053241 512905543662054235449961761129912643539514261930778780575803129094809217414988287315500794889688083726470478732454337581 

20170624, 01:00  #135  
Aug 2006
3·1,993 Posts 
Quote:


20170624, 12:39  #136  
Feb 2017
Nowhere
2^{2}×3^{2}×139 Posts 
Quote:
Besides, I have said myself that your "method" is likely to stumble across factors less than 100. You haven't tried your "method" on any numbers large enough to be of interest WRT whether your "new factoring method" has any merit. 

20170624, 12:55  #137  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
Quote:
The algorithm is described here. https://math.stackexchange.com/quest...thataddupto To answer your second question, there is no quick way to tell if a number does not admit a representation as a sum of two squares. To find out, one has to check every value of the parameter k in the discriminant until we run out of values to check before we can say the number does not admit a representation as a sum of two squares. And this may take a long time, maybe even longer than factorization. 

20170624, 13:19  #138  
Feb 2017
Nowhere
2^{2}×3^{2}×139 Posts 
Quote:
Quote:
8*N  4*(k + 1)^2 = s^2, or 8*N = s^2 + 4*(k + 1)^2. In other words, the condition that there is a solution, is that 8*N be a sum of two squares. And how do we actually find them? Quote:
Last fiddled with by Dr Sardonicus on 20170624 at 13:28 

20170624, 13:38  #139 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
This proves that s is even, because in general . If s=2*t then s^2= 4(t^2). We can remove the factor of 4 from each side, reducing it to 2N=t^2+(k+1)^2 so the fact that 8N has to be of this form, proves 2N has to a sum of two squares.
Last fiddled with by science_man_88 on 20170624 at 13:42 
20170624, 13:46  #140  
Feb 2017
Nowhere
2^{2}×3^{2}×139 Posts 
Quote:
I think that others have already shown that your "method" is extremely unlikely to work in a reasonable amount of time with numbers of any size. The fact that you have steadfastly refused to apply your method to challenge numbers of any size lends credence to this assessment. Besides, in your above demand, you appear to be admitting that your method is only guaranteed to work if you are allowed to combine the 4 numbers without limitation  which, as far as I can tell, could include substituting a, b, c, and d into any polynomial in 4 variables with integer coefficients. If you're saying that that's what it would take to guarantee that your "method" would produce a factor, then it is as I previously described it: blazing away with a scattergun, desperately hoping that some of the buckshot will hit. 

20170624, 13:59  #141 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
The answer Will Jagy gave in the thread on the statckexchange site proved that writing 2N as a sum of two squares is equivalent to writing N as a sum of two squares. And he is one of the most knowledgeable mathematician I know. He called it a bijection between the two methods.

20170624, 14:38  #142  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20170624 at 15:08 

20170624, 18:57  #143  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9572_{10} Posts 
Quote:
Doesn't it sound like a question, rather than an answer? Why yes! It is a question, and the answer is 'No.' There is no algorithm in that thread. Try again. 

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