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Homesums of two squares

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# sums of two squares

###### Theorem.

The set of the sums of two squares of integers is closed under multiplication^{}; in fact we have the identical equation

$\displaystyle(a^{2}\!+\!b^{2})(c^{2}\!+\!d^{2})\;=\;(ac\!-\!bd)^{2}\!+\!(ad\!+% \!bc)^{2}.$ | (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 $8^{2}\!+\!1^{2}$.

Related:

EulerFourSquareIdentity, TheoremsOnSumsOfSquares, DifferenceOfSquares

Synonym:

Diophantus' identity, Brahmagupta's identity, Fibonacci's identity

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

11A67*no label found*11E25

*no label found*

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