KGS math club/solution 4 1

(a²+b²)(c²+d²) = (ac-bd)²+(ad+bd)²

and

(a1²+a2²+a3²+a4²)(b1²+b2²+b3²+b4²)= (a1b1 - a2b2 - a3b3 - a4b4)² + (a1b2 + a2b1 + a3b4 - a4b3)² + (a1b3 - a2b4 + a3b1 + a4b2)² + (a1b4 + a2b3 - a3b2 + a4b1)²

Remarks :

- It's in fact nothing more than 1) the norms of the complex numbers z=a+ib, z'=c+id, and the product zz' 2) the norms of the quaternion numbers z=a1+a2i+a3j+a4k, z'=b1+b2i+b3j+b4k, and the product zz'

- Every non negative integer is a sum of four squares, so the second part of the problem makes little sense.

- There is another identity with 8 squares : the product of two sums of 8 squares is a sum of 8 squares (we just have to use the same technic, with octonion numbers). However there is no identity with 16 squares or more...