proof of Euler four-square identity

Using Lagrange’s identity, we have

${\left({\displaystyle \sum _{k=1}^4}{x}_k{y}_k\right)}^2$(1)

We group the six squares into 3 groups of two squares and rewrite:

${(x_1{y}_2-x_2{y}_1)}^2+{(x_3{y}_4-x_4{y}_3)}^2$(2)

$=$${((x_1{y}_2-x_2{y}_1)+(x_3{y}_4-x_4{y}_3))}^2-2((x_1{y}_2-x_2{y}_1)(x_3{y}_4-x_4{y}_3))$(3)

${(x_1{y}_3-x_3{y}_1)}^2+{(x_2{y}_4-x_4{y}_2)}^2$

$=$${((x_1{y}_3-x_3{y}_1)-(x_2{y}_4-x_4{y}_2))}^2+2(x_1{y}_3-x_3{y}_1)(x_2{y}_4-x_4{y}_2)$(4)

${(x_1{y}_4-x_4{y}_1)}^2+{(x_2{y}_3-x_3{y}_2)}^2$

$=$${((x_1{y}_4-x_4{y}_1)+(x_2{y}_3-x_3{y}_2))}^2-2(x_1{y}_4-x_4{y}_1)(x_2{y}_3-x_3{y}_2)\text{.}$(5)

Using

$-2((x_1{y}_2-x_2{y}_1)(x_3{y}_4-x_4{y}_3))$$+2(x_1{y}_3-x_3{y}_1)(x_2{y}_4-x_4{y}_2)$(6)

$-2(x_1{y}_4-x_4{y}_1)(x_2{y}_3-x_3{y}_2)$$=0$

we get

$=((x_1{y}_2-x_2{y}_1)$$+(x_3{y}_4-x_4{y}_3)){}^2$(7)

$+{((x_1{y}_3-x_3{y}_1)-(x_2{y}_4-x_4{y}_2))}^2$(8)

$+{((x_1{y}_4-x_4{y}_1)+(x_2{y}_3-x_3{y}_2))}^2$

by adding equations 2-4. We put the result of equation 7 into1 and get

${\left({\displaystyle \sum _{k=1}^4}{x}_k{y}_k\right)}^2$(9)

$=\left({\displaystyle \sum _{k=1}^4}{x}_k^2\right)\left({\displaystyle \sum _{k=1}^4}{y}_k^2\right)$$-({(x_1{y}_2-x_2{y}_1+x_3{y}_4-x_4{y}_3)}^2$

$-{(x_1{y}_3-x_3{y}_1+x_4{y}_2-x_2{y}_4)}^2$$-{(x_1{y}_4-x_4{y}_1+x_2{y}_3-x_3{y}_2)}^2$

which is equivalent to the claimed identity.