Talk:Euler's four-square identity

Quaternion multiplication

I modified the specific 4-square identity so that it has an all plus term; it now corresponds to the quaternion multiplication $b{\bar {a}},$, rather than the former $ba.$ If someone wants to change it to $a{\bar {b}},$ I have no objection, but it seems important to me that one term is $\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}+a_{4}b_{4}\right)^{2}.$ — Arthur Rubin (talk) 02:45, 2 July 2014 (UTC)