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20170627, 23:34  #177  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
Quote:


20170628, 00:30  #178 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
It also means, that looking for the partitions where all the number's are the same, is enough to get a factor though.

20170628, 12:03  #179 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20170628, 12:35  #180  
Feb 2017
Nowhere
7·733 Posts 
Quote:
Instead of looking at summands n^2, why not use summands 2^n instead? I absolutely guarantee, if N is odd and composite, and you take 1, 2, 4, ... 2^r, with 2^r < sqrt(N) < 2^(r+1), so r < log(N)/(2*log(2)), then at least one of the sums of these powers of 2 will be equal to a factor of N! I can even spot you that one of the summands must be equal to 1! Last fiddled with by Dr Sardonicus on 20170628 at 12:39 

20170628, 13:13  #181  
Aug 2006
3·1,993 Posts 
Quote:


20170628, 13:25  #182  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
Quote:
Because we are dealing with a big number, the squares (a,b,c,d) themselves will have many 4sq reps. The idea of my comment is to limit the search for factors to 4 sub expansions, that is the 1st 4sq rep of a,b,c,d. 

20170628, 13:36  #183 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
can't divide those big number's by 2 ? after all (2a)^2= 4(a^2) edit :to prove what you said, most of what you have to do is prove that, any of the squares, once expanded form a lower pair of a Pythagorean triple.
Last fiddled with by science_man_88 on 20170628 at 13:50 
20170628, 19:46  #184 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
not necessarily a lower pair of Pythagorean triple. 8^2=4^2+4^2+4^2+4^2 but 4^2=2^2+2^2+2^2+2^2. We may get some Pythagorean triplets but not as a general rule.

20170628, 20:41  #185 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20170629, 12:21  #186 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
It does. Expanding the squares of a given 4sq rep helps as a general rule. Look at the simple example of N=7*13=91, the first 4sq rep is (5,5,5,4). There is not a single combination of squares that gives a factor. But there is at least one combination of nonsquare that gives a factor, 5+5+4=14=2*7. There is also a mixed combination that gives a factor, 5^2 + 5 + 5 +4 = 39 =3*13.
Now look at what happens when we expand the squares (5,5,5,4). 5^2=(4,2,2,1) 4^2=(2,2,2,2) combining these squares, it is easy to see that: 4^2 + 2^2 + 1^2 = 21 = 3*7 2^2 + 2^2 + 2^2 + 1^2 = 13 2^2 + 1^2 + 1^2 + 1^2 = 7 4^2 + 2^2 + 2^2 + 2^2 = 28 = 4*7 4^2 + 4^2 + 1^2 + 1^2 + 1^2 = 35 = 5*7 .... What we don't know is how many levels of expansions are needed. If we consider the original 4sq rep at the top of a pyramid, the successive expansions of squares of that original 4sq rep will take us all the way down to a^2=1^2+1^2+...+1^2 and the same for b,c,d. My guess is that it is very hard to prove that the number of expansions of individual squares of the original 4sq rep is much smaller than the one required if one has to go all the way down to a^2=1^2+1^2+...+1^2. The math is not simple and I don't think the answer is the same for all numbers. Last fiddled with by mahbel on 20170629 at 12:23 Reason: fix a typo 
20170629, 12:36  #187  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
so 1^2+3^2+4^2+5^2 =1^2+5^2+5^2+0^2 because 3^2+4^2=5^2. 

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