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 $$b a.$$ 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)