Quote:
Originally Posted by mahbel
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.

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.