Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it well-known that there are an infinite number of Euler-type 4-square identities? Proof:

$\begin{align} &{(x_1^2+ x_2^2+ x_3^2+ x_4^2) (y_1^2+ y_2^2+ y_3^2+ y_4^2)\,=\,z_1^2+ z_2^2+ z_3^2+ z_4^2}\\ &\text{where,}\\ &{z_1 \,=\, x_1y_1+a_1y_2+ a_2y_3+ a_3y_4}\\ &{z_2 \,=\, x_2y_1+b_1y_2+ b_2y_3+ b_3y_4}\\ &{z_3 \,=\, x_3y_1+c_1y_2+ c_2y_3+ c_3y_4}\\ &{z_4 \,=\, x_4y_1+x_3y_2-x_2y_3+x_1y_4}\\ &\text{and,}\\ &{a_1 \,=\, -k(x_1x_3-nx_2)\,+\,x_2,\;\; b_1 \,=\, -k(x_2x_3+nx_1)\,-\,x_1,\;\; c_1 \,=\, k(x_1^2+x_2^2)\,-\,x_4 }\\ &{a_2 \,=\, k(x_1x_2+nx_3)\,+\,x_3,\;\;\;\; b_2 \,=\, -k(x_1^2+x_3^2)\,+\,x_4,\;\;\;\;\;\;\;\; c_2 \,=\, k(x_2x_3-nx_1)\,-\,x_1 }\\ &{a_3 \,=\, k(x_2^2+x_3^2)\,-\,x_4,\;\;\;\;\;\;\;\;\; b_3 \,=\, -k(x_1x_2-nx_3)\,+\,x_3,\;\; c_3 \,=\, -k(x_1x_3+nx_2)\,-\,x_2}\\ &{k = \frac{2(x_4-n)}{x_1^2+x_2^2+x_3^2+n^2}}\\ \end{align}$

for arbitrary n. Of course, for $n = x_4$, then $k = 0$, all the $z_i$ become bilinear, hence the Euler 4-square identity is a special case of this family. In fact, there is an even larger 8-square family. Is it also already known there is an infinite number of Degen-type 8-square identities? More details here.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.