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2017-06-29, 12:21   #186
mahbel

Feb 2017
Kitchener, Ontario

22·3·5 Posts

Quote:
 Originally Posted by science_man_88 then breaking them down doesn't help in general ...
It does. Expanding the squares of a given 4-sq rep helps as a general rule. Look at the simple example of N=7*13=91, the first 4-sq 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 non-square 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 4-sq rep at the top of a pyramid, the successive expansions of squares of that original 4-sq 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 4-sq 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 2017-06-29 at 12:23 Reason: fix a typo