From: ikastan@sol.uucp (ilias kastanas 08-14-90)
Newsgroups: sci.math
Subject: Re: Numbers problem (sums of squares)
Date: 19 Aug 1998 16:33:39 GMT

In article <1998081902273700.WAA28067@ladder03.news.aol.com>,
KRamsay <kramsay@aol.com> wrote:
@In article <Pine.SUN.3.96.980818021105.29192A-100000@stlawrence.maths>, 
@Jan Kristian Haugland <haugland@maths.ox.ac.uk> writes:
@|Homework or not, here goes:
@|
@|   (a+b)^2 + (a-b)^2 = 2(a^2 + b^2)
@
@We should point out that this is a special case of the multiplication
@law (a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2 which shows the 
@product of two sums-of-two-squares is also one. It can be shown
@as a consequence of the factorization a^2+b^2=(a+bi)(a-bi). There
@is a similar formula for sums of four squares, based on quaternions.



	The story goes further.   There is a similar formula for sums
of 8 squares.  (One way to get it is via octonions; Norm(ab) = Norm(a)Norm(b)
... just like the two cases above.   I won't write it out).

	What about values other than  n = (1), 2, 4, 8?


	It depends on what we mean.   We've found identities of the form

Sum(x_i ^2) * Sum(y_i ^2) = Sum(f_i ^2)    (i from  1   to  n)

with  f_i in  Z[x_i..., y_i...]   (polynomials with integer coefficients),
and bilinear.   If _that_ is what we want, for other  n ...  too bad; there
aren't any!


	Suppose however we relax the requirement, and only ask that  "sum
of  n  squares  *  sum of  n  squares   be a sum of  n  squares"... never
mind "how"!    There is a precise answer then:

	The condition holds, in all fields  K,  if  n = 2^k.

And this is tight, in the sense: it is only for n = power of 2  that the
condition can hold in all fields.




						Ilias