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 into 1 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.

Title	proof of Euler four-square identity
Canonical name	ProofOfEulerFoursquareIdentity
Date of creation	2013-03-22 13:18:10
Last modified on	2013-03-22 13:18:10
Owner	Thomas Heye (1234)
Last modified by	Thomas Heye (1234)
Numerical id	7
Author	Thomas Heye (1234)
Entry type	Proof
Classification	msc 13A99