# Talk:Euler's four-square identity

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)