20170623, 13:26  #122  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
Quote:
I don't know if a faster way to generate sums of 4 squares can be found. The online calculator I used is very slow when fed large numbers. There is this link on another way to calculate the sum of squares but I couldn't find an online calculator based on the algorithm described in the link. I have no idea if it is a faster algorithm than the one we are using. https://cs.stackexchange.com/questio...thatsumton In looking for ways to improve the method, one should maybe consider dropping the requirement that 4sq reps be used. After all, there is no theoretical justification (I know of) that only sums of 4 squares can be used to factor integers. One may consider using sums of 5, 6 or more squares to factor. And here again, there is no theoretical justification that says sums of 5, 6,,,squares cannot factor integers. If the requirement to use a sum of 4 squares is dropped, then it is possible to quickly find a representation with a variable number of squares by reducing the number N with the square root function with a, the first square calculated as before. Once we have a representation of a sum of n squares, we test it as usual by calculating the gcd's. If no factor is found, then instead of calculating another representation, we take one square from the previous sum, say b, and expand it into a sum of m squares. The b square will be replaced by m squares in the original sum. The obvious drawback is that the number of combinations to test will grow rapidly if no factor is found quickly. I have not had a chance to test these ideas thoroughly like it was done with the sum of 4 squares method so at this point nothing can be said. The other way to improve the method is to do a serious mathematical analysis of what is involved in factoring when using a sum of 4 squares. One can try to understand why some 4sq reps lead to a factor and some don't. There is also the need to understand the distribution of the good representations (the ones that give a factor) within the set of a 4sq sum of a given integer but also the order in which they appear. A major improvement will be to find a way to make the good representations appear at the beginning of the sum of 4 squares. Sadly, all this is beyond what I can do. Last fiddled with by mahbel on 20170623 at 13:27 Reason: correct spelling. 

20170623, 13:50  #123  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20170623 at 13:51 

20170623, 14:12  #124  
Jun 2003
3·17·101 Posts 
Quote:
Quote:
You haven't tested _any_ of your ideas thoroughly. You have neither the mathematical ability to theoretically evaluate your methods, nor the computational knowhow to empirically evaluate them. People have given hints about how to fix the latter, but you're hell bent on ignoring it. 

20170623, 15:42  #125  
Feb 2017
Nowhere
2^{2}·3^{2}·139 Posts 
In remarkes addressed to the OP:
Quote:
From the web site listed under the "Miscellaneous" heading, I offer Quote:
N = 77, 1457, 3901, 4661, 6557, 7597, 8549, and 9869 fill the bill. Let the OP try to use his "method" to factor any of these which are larger than 77... Last fiddled with by Dr Sardonicus on 20170623 at 15:55 

20170623, 16:14  #126  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,393 Posts 
Quote:
______________________ *meaning that this generator will not run out of variants. So the factorization is solved by "reaching into the bingo cage" ridiculously many times. 

20170623, 19:33  #127  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
Quote:
factor: 36+11=47 1457 = 31*47 = 27^2 + 20^2 +18^2 +2^2 factor: 27+20=47 1457 = 31*47 = 30^2 + 21^2 + 10^2 + 4^2 factor: 21+10=31 I don't know if you need to see more ways to get a factor by combining nonsquares (a,b,c,d). There may be a lot more. In fact, I quickly checked that there are more (30,20,11,6). I did not check if no combinations of squares could provide a factor. I also did not check if any mixed combinations ( squares + non squares) could provide a factor. If you can give us a rational justification for why the above calculated factors must be rejected, then we will be very grateful. Last fiddled with by mahbel on 20170623 at 19:41 Reason: added text 

20170623, 19:53  #128  
Feb 2017
Nowhere
11614_{8} Posts 
Quote:
It might be interesting to see how quickly this "circular method" works  that is, start with a factor p, and see how long it takes to find a and b with a + b = p, and N  a^2  b^2 = c^2 + d^2. 

20170623, 20:38  #129  
Aug 2006
3·1,993 Posts 
Quote:


20170623, 23:38  #130  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
Quote:
a+b, a+c, a+d, b+c, b+d, a+b+c, a+b+d, b+c+d and you calculate the gcd of these combinations. There is a really simple way to see if I somehow cheated and invented these combinations that sum up to a factor. You just do it yourself. I can tell you that I used the online 4sq reps calculator for which I provided the link before. Now, if you want to impose limitations on how one can go about combining 4 different numbers in different ways, then please post your do's and dont's. 

20170623, 23:42  #131  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
Quote:


20170624, 00:23  #132 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
Expanding Fermat's factorization method:
a+b  a^n + b^n For all positive odd integers n. Similarly, all positive even integers n, yield imaginary integer factors. However, while true it has little practical value for large semi primes, due to "exponential" growth of powers. Last fiddled with by a1call on 20170624 at 00:31 
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