# proof of Euler four-square identity

 $\displaystyle\left(\sum_{k=1}^{4}x_{k}y_{k}\right)^{2}$ $\displaystyle=\left(\sum_{k=1}^{4}x_{k}^{2}\right)\left(\sum_{k=1}^{4}y_{k}^{2% }\right)-\sum_{1\leq k (1)

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

 $\displaystyle(x_{1}y_{2}-x_{2}y_{1})^{2}+(x_{3}y_{4}-x_{4}y_{3})^{2}$ (2) $\displaystyle=$ $\displaystyle((x_{1}y_{2}-x_{2}y_{1})+(x_{3}y_{4}-x_{4}y_{3}))^{2}-2((x_{1}y_{% 2}-x_{2}y_{1})(x_{3}y_{4}-x_{4}y_{3}))$ (3) $\displaystyle(x_{1}y_{3}-x_{3}y_{1})^{2}+(x_{2}y_{4}-x_{4}y_{2})^{2}$ $\displaystyle=$ $\displaystyle((x_{1}y_{3}-x_{3}y_{1})-(x_{2}y_{4}-x_{4}y_{2}))^{2}+2(x_{1}y_{3% }-x_{3}y_{1})(x_{2}y_{4}-x_{4}y_{2})$ (4) $\displaystyle(x_{1}y_{4}-x_{4}y_{1})^{2}+(x_{2}y_{3}-x_{3}y_{2})^{2}$ $\displaystyle=$ $\displaystyle((x_{1}y_{4}-x_{4}y_{1})+(x_{2}y_{3}-x_{3}y_{2}))^{2}-2(x_{1}y_{4% }-x_{4}y_{1})(x_{2}y_{3}-x_{3}y_{2})\mbox{.}$ (5)

Using

 $\displaystyle-2((x_{1}y_{2}-x_{2}y_{1})(x_{3}y_{4}-x_{4}y_{3}))$ $\displaystyle+2(x_{1}y_{3}-x_{3}y_{1})(x_{2}y_{4}-x_{4}y_{2})$ (6) $\displaystyle-2(x_{1}y_{4}-x_{4}y_{1})(x_{2}y_{3}-x_{3}y_{2})$ $\displaystyle=0$

we get

 $\displaystyle\sum_{1\leq k $\displaystyle=((x_{1}y_{2}-x_{2}y_{1})$ $\displaystyle+(x_{3}y_{4}-x_{4}y_{3}))^{2}$ (7) $\displaystyle+((x_{1}y_{3}-x_{3}y_{1})-(x_{2}y_{4}-x_{4}y_{2}))^{2}$ (8) $\displaystyle+((x_{1}y_{4}-x_{4}y_{1})+(x_{2}y_{3}-x_{3}y_{2}))^{2}$

by adding equations 2-4. We put the result of equation 7 into 1 and get

 $\displaystyle\left(\sum_{k=1}^{4}x_{k}y_{k}\right)^{2}$ (9) $\displaystyle=\left(\sum_{k=1}^{4}x_{k}^{2}\right)\left(\sum_{k=1}^{4}y_{k}^{2% }\right)$ $\displaystyle-((x_{1}y_{2}-x_{2}y_{1}+x_{3}y_{4}-x_{4}y_{3})^{2}$ $\displaystyle-(x_{1}y_{3}-x_{3}y_{1}+x_{4}y_{2}-x_{2}y_{4})^{2}$ $\displaystyle-(x_{1}y_{4}-x_{4}y_{1}+x_{2}y_{3}-x_{3}y_{2})^{2}$
Title proof of Euler four-square identity ProofOfEulerFoursquareIdentity 2013-03-22 13:18:10 2013-03-22 13:18:10 Thomas Heye (1234) Thomas Heye (1234) 7 Thomas Heye (1234) Proof msc 13A99