Degen's eight-square identity

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

$${\displaystyle {\begin{aligned}&\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}+b_{5}^{2}+b_{6}^{2}+b_{7}^{2}+b_{8}^{2}\right)=\\[1ex]&\quad \left(a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}-a_{4}b_{4}-a_{5}b_{5}-a_{6}b_{6}-a_{7}b_{7}-a_{8}b_{8}\right)^{2}+\\&\quad \left(a_{1}b_{2}+a_{2}b_{1}+a_{3}b_{4}-a_{4}b_{3}+a_{5}b_{6}-a_{6}b_{5}-a_{7}b_{8}+a_{8}b_{7}\right)^{2}+\\&\quad \left(a_{1}b_{3}-a_{2}b_{4}+a_{3}b_{1}+a_{4}b_{2}+a_{5}b_{7}+a_{6}b_{8}-a_{7}b_{5}-a_{8}b_{6}\right)^{2}+\\&\quad \left(a_{1}b_{4}+a_{2}b_{3}-a_{3}b_{2}+a_{4}b_{1}+a_{5}b_{8}-a_{6}b_{7}+a_{7}b_{6}-a_{8}b_{5}\right)^{2}+\\&\quad \left(a_{1}b_{5}-a_{2}b_{6}-a_{3}b_{7}-a_{4}b_{8}+a_{5}b_{1}+a_{6}b_{2}+a_{7}b_{3}+a_{8}b_{4}\right)^{2}+\\&\quad \left(a_{1}b_{6}+a_{2}b_{5}-a_{3}b_{8}+a_{4}b_{7}-a_{5}b_{2}+a_{6}b_{1}-a_{7}b_{4}+a_{8}b_{3}\right)^{2}+\\&\quad \left(a_{1}b_{7}+a_{2}b_{8}+a_{3}b_{5}-a_{4}b_{6}-a_{5}b_{3}+a_{6}b_{4}+a_{7}b_{1}-a_{8}b_{2}\right)^{2}+\\&\quad \left(a_{1}b_{8}-a_{2}b_{7}+a_{3}b_{6}+a_{4}b_{5}-a_{5}b_{4}-a_{6}b_{3}+a_{7}b_{2}+a_{8}b_{1}\right)^{2}\end{aligned}}}$$

First discovered by

of product of two octonions equals the product of their norms: ${\displaystyle \left\|ab\right\|=\left\|a\right\|\left\|b\right\|}$. Similar statements are true for quaternions (

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

$${\displaystyle {\begin{aligned}&\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\right)=\\&\quad \left(a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}-a_{4}b_{4}\right)^{2}+\\&\quad \left(a_{1}b_{2}+a_{2}b_{1}+a_{3}b_{4}-a_{4}b_{3}\right)^{2}+\\&\quad \left(a_{1}b_{3}-a_{2}b_{4}+a_{3}b_{1}+a_{4}b_{2}\right)^{2}+\\&\quad \left(a_{1}b_{4}+a_{2}b_{3}-a_{3}b_{2}+a_{4}b_{1}\right)^{2}\end{aligned}}}$$

and similarly for the other three quadrants.

Comment: The proof of the eight-square identity is by algebraic evaluation. The eight-square identity can be written in the form of a product of two inner products of 8-dimensional vectors, yielding again an inner product of 8-dimensional vectors: (a·a)(b·b) = (a×b)·(a×b). This defines the octonion multiplication rule a×b, which reflects Degen's 8-square identity and the mathematics of octonions.

By

${\displaystyle z_{i}}$

${\displaystyle x_{i},y_{i}}$

$${\displaystyle \left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}+y_{5}^{2}+y_{6}^{2}+y_{7}^{2}+y_{8}^{2}\right)=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}+z_{5}^{2}+z_{6}^{2}+z_{7}^{2}+z_{8}^{2}}$$

where,

$${\displaystyle {\begin{aligned}z_{1}&=x_{1}y_{1}-x_{2}y_{2}-x_{3}y_{3}-x_{4}y_{4}+u_{1}y_{5}-u_{2}y_{6}-u_{3}y_{7}-u_{4}y_{8}\\z_{2}&=x_{2}y_{1}+x_{1}y_{2}+x_{4}y_{3}-x_{3}y_{4}+u_{2}y_{5}+u_{1}y_{6}+u_{4}y_{7}-u_{3}y_{8}\\z_{3}&=x_{3}y_{1}-x_{4}y_{2}+x_{1}y_{3}+x_{2}y_{4}+u_{3}y_{5}-u_{4}y_{6}+u_{1}y_{7}+u_{2}y_{8}\\z_{4}&=x_{4}y_{1}+x_{3}y_{2}-x_{2}y_{3}+x_{1}y_{4}+u_{4}y_{5}+u_{3}y_{6}-u_{2}y_{7}+u_{1}y_{8}\\z_{5}&=x_{5}y_{1}-x_{6}y_{2}-x_{7}y_{3}-x_{8}y_{4}+x_{1}y_{5}-x_{2}y_{6}-x_{3}y_{7}-x_{4}y_{8}\\z_{6}&=x_{6}y_{1}+x_{5}y_{2}+x_{8}y_{3}-x_{7}y_{4}+x_{2}y_{5}+x_{1}y_{6}+x_{4}y_{7}-x_{3}y_{8}\\z_{7}&=x_{7}y_{1}-x_{8}y_{2}+x_{5}y_{3}+x_{6}y_{4}+x_{3}y_{5}-x_{4}y_{6}+x_{1}y_{7}+x_{2}y_{8}\\z_{8}&=x_{8}y_{1}+x_{7}y_{2}-x_{6}y_{3}+x_{5}y_{4}+x_{4}y_{5}+x_{3}y_{6}-x_{2}y_{7}+x_{1}y_{8}\end{aligned}}}$$

and,

$${\displaystyle {\begin{aligned}u_{1}&={\frac {(ax_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2})x_{5}-2x_{1}(bx_{1}x_{5}+x_{2}x_{6}+x_{3}x_{7}+x_{4}x_{8})}{c}}\\u_{2}&={\frac {(x_{1}^{2}+ax_{2}^{2}+x_{3}^{2}+x_{4}^{2})x_{6}-2x_{2}(x_{1}x_{5}+bx_{2}x_{6}+x_{3}x_{7}+x_{4}x_{8})}{c}}\\u_{3}&={\frac {(x_{1}^{2}+x_{2}^{2}+ax_{3}^{2}+x_{4}^{2})x_{7}-2x_{3}(x_{1}x_{5}+x_{2}x_{6}+bx_{3}x_{7}+x_{4}x_{8})}{c}}\\u_{4}&={\frac {(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+ax_{4}^{2})x_{8}-2x_{4}(x_{1}x_{5}+x_{2}x_{6}+x_{3}x_{7}+bx_{4}x_{8})}{c}}\end{aligned}}}$$

with,

$${\displaystyle a=-1,\;\;b=0,\;\;c=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$$

Incidentally, the ${\displaystyle u_{i}}$ obey the identity,

$${\displaystyle u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}=x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2}}$$

See also

• Pfister's sixteen-square identity
• Cayley–Dickson construction
• Hypercomplex number
• Latin square

External links

• Degen's eight-square identity on MathWorld
• The Degen–Graves–Cayley Eight-Square Identity
• Pfister's 16-Square Identity Archived 2016-09-17 at the Wayback Machine