View Single Post 2017-06-29, 12:36   #187
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts Quote:
 Originally Posted by mahbel 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.
you said 4 square representations to another not a mixed representation. to get back squares you need a sum of squares that sum to a square the simplest form is a pythagorean triple.

so 1^2+3^2+4^2+5^2 =1^2+5^2+5^2+0^2 because 3^2+4^2=5^2.  