# sums of two squares

###### Theorem.
 $\displaystyle(a^{2}\!+\!b^{2})(c^{2}\!+\!d^{2})\;=\;(ac\!-\!bd)^{2}\!+\!(ad\!+% \!bc)^{2}.$ (1)

This was presented by Leonardo Fibonacci in 1225 (in Liber quadratorum), but was known also by Brahmagupta and already by Diophantus of Alexandria (III book of his Arithmetica).

The proof of the equation may utilize Gaussian integers  as follows:

 $\displaystyle(a^{2}\!+\!b^{2})(c^{2}\!+\!d^{2})$ $\displaystyle\;=\;(a\!+\!ib)(a\!-\!ib)(c\!+\!id)(c\!-\!id)$ $\displaystyle\;=\;(a\!+\!ib)(c\!+\!id)(a\!-\!ib)(c\!-\!id)$ $\displaystyle\;=\;[(ac\!-\!bd)\!+\!i(ad\!+\!bc)][(ac\!-\!bd)\!-\!i(ad\!+\!bc)]$ $\displaystyle\;=\;(ac\!-\!bd)^{2}\!+\!(ad\!+\!bc)^{2}$

Note 1.  The equation (1) is the special case  $n=2$  of Lagrange’s identity.

Note 2.  Similarly as (1), one can derive the identity

 $\displaystyle(a^{2}\!+\!b^{2})(c^{2}\!+\!d^{2})\;=\;(ac\!+\!bd)^{2}\!+\!(ad\!-% \!bc)^{2}.$ (2)

Thus in most cases, we can get two different nontrivial sum forms (i.e. without a zero addend) for a given product of two sums of squares.  For example, the product

 $65=5\!\cdot\!13=(2^{2}\!+\!1^{2})(3^{2}\!+\!2^{2})$

attains the two forms $4^{2}\!+\!7^{2}$ and $8^{2}\!+\!1^{2}$.

 Title sums of two squares Canonical name SumsOfTwoSquares Date of creation 2013-11-19 16:28:21 Last modified on 2013-11-19 16:28:21 Owner pahio (2872) Last modified by pahio (2872) Numerical id 33 Author pahio (2872) Entry type Theorem  Classification msc 11A67 Classification msc 11E25 Synonym Diophantus’ identity Synonym Brahmagupta’s identity Synonym Fibonacci’s identity Related topic EulerFourSquareIdentity Related topic TheoremsOnSumsOfSquares Related topic DifferenceOfSquares