# Degen's eight-square identity

In mathematics, Degen's eight-square identity establishes that the product of two numbers, each of which being a sum of eight squares, is itself the sum of eight squares. Namely:

$(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2+a_7^2+a_8^2)(b_1^2+b_2^2+b_3^2+b_4^2+b_5^2+b_6^2+b_7^2+b_8^2)=\,$
$(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4 - a_5b_5 - a_6b_6 - a_7b_7 - a_8b_8)^2+\,$
$(a_1b_2 + a_2b_1 + a_3b_4 - a_4b_3 + a_5b_6 - a_6b_5 - a_7b_8 + a_8b_7)^2+\,$
$(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2 + a_5b_7 + a_6b_8 - a_7b_5 - a_8b_6)^2+\,$
$(a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1 + a_5b_8 - a_6b_7 + a_7b_6 - a_8b_5)^2+\,$
$(a_1b_5 - a_2b_6 - a_3b_7 - a_4b_8 + a_5b_1 + a_6b_2 + a_7b_3 + a_8b_4)^2+\,$
$(a_1b_6 + a_2b_5 - a_3b_8 + a_4b_7 - a_5b_2 + a_6b_1 - a_7b_4 + a_8b_3)^2+\,$
$(a_1b_7 + a_2b_8 + a_3b_5 - a_4b_6 - a_5b_3 + a_6b_4 + a_7b_1 - a_8b_2)^2+\,$
$(a_1b_8 - a_2b_7 + a_3b_6 + a_4b_5 - a_5b_4 - a_6b_3 + a_7b_2 + a_8b_1)^2\,$

First discovered by Ferdinand Degen (Danish) around 1818, the identity was independently rediscovered by John Thomas Graves (1843) and Arthur Cayley (1845). The latter two derived it while working on an extension of quaternions called octonions. In algebraic terms the identity means that the norm of product of two octonions equals the product of their norms: $\|ab\| = \|a\|\|b\|$. Similar statements are true for quaternions (Euler's four-square identity), complex numbers (the Brahmagupta–Fibonacci two-square identity) and real numbers. In 1898 Adolf Hurwitz proved that there is no similar bilinear identity for 16 squares (sedenions) or any other number of squares except for 1,2,4, and 8. However, in the 1960s, H. Zassenhaus, W. Eichhorn, and A. Pfister (independently) showed there can be a non-bilinear identity for 16 squares.

Note that each quadrant reduces to a version of Euler's four-square identity:

$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)=\,$
$(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4)^2+\,$
$(a_1b_2 + a_2b_1 + a_3b_4 - a_4b_3)^2+\,$
$(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2)^2+\,$
$(a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1)^2\,$

and similarly for the other three quadrants. By Pfister's theorem, a different sort of eight-square identity can be given where the $z_i$, introduced below, are non-bilinear and merely rational functions of the $x_i, y_i$. Thus,

$(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2+x_8^2)(y_1^2+y_2^2+y_3^2+y_4^2+y_5^2+y_6^2+y_7^2+y_8^2) = z_1^2+z_2^2+z_3^2+z_4^2+z_5^2+z_6^2+z_7^2+z_8^2$

where,

$z_1 = x_1 y_1 - x_2 y_2 - x_3 y_3 - x_4 y_4 + u_1 y_5 - u_2 y_6 - u_3 y_7 - u_4 y_8$
$z_2 = x_2 y_1 + x_1 y_2 + x_4 y_3 - x_3 y_4 + u_2 y_5 + u_1 y_6 + u_4 y_7 - u_3 y_8$
$z_3 = x_3 y_1 - x_4 y_2 + x_1 y_3 + x_2 y_4 + u_3 y_5 - u_4 y_6 + u_1 y_7 + u_2 y_8$
$z_4 = x_4 y_1 + x_3 y_2 - x_2 y_3 + x_1 y_4 + u_4 y_5 + u_3 y_6 - u_2 y_7 + u_1 y_8$
$z_5 = x_5 y_1 - x_6 y_2 - x_7 y_3 - x_8 y_4 + x_1 y_5 - x_2 y_6 - x_3 y_7 - x_4 y_8$
$z_6 = x_6 y_1 + x_5 y_2 + x_8 y_3 - x_7 y_4 + x_2 y_5 + x_1 y_6 + x_4 y_7 - x_3 y_8$
$z_7 = x_7 y_1 - x_8 y_2 + x_5 y_3 + x_6 y_4 + x_3 y_5 - x_4 y_6 + x_1 y_7 + x_2 y_8$
$z_8 = x_8 y_1 + x_7 y_2 - x_6 y_3 + x_5 y_4 + x_4 y_5 + x_3 y_6 - x_2 y_7 + x_1 y_8$

and,

$u_1 = \frac{(ax_1^2+x_2^2+x_3^2+x_4^2)x_5 - 2x_1(bx_1 x_5 + x_2 x_6+ x_3 x_7+ x_4 x_8)}{c}$
$u_2 = \frac{(x_1^2+ax_2^2+x_3^2+x_4^2)x_6 - 2x_2(x_1 x_5 + bx_2 x_6+ x_3 x_7+ x_4 x_8)}{c}$
$u_3 = \frac{(x_1^2+x_2^2+ax_3^2+x_4^2)x_7 - 2x_3(x_1 x_5 + x_2 x_6+ bx_3 x_7+ x_4 x_8)}{c}$
$u_4 = \frac{(x_1^2+x_2^2+x_3^2+ax_4^2)x_8 - 2x_4(x_1 x_5 + x_2 x_6+ x_3 x_7+ bx_4 x_8)}{c}$

with,

$a=-1,\;\; b=0,\;\; c=x_1^2+x_2^2+x_3^2+x_4^2$

Incidentally, the $u_i$ obey the identity,

$u_1^2+u_2^2+u_3^2+u_4^2 = x_5^2+x_6^2+x_7^2+x_8^2$