Brahmagupta-Fibonacci Identity

Theorem

Let $a, b, c, d$ be numbers.

Then:

	\(\displaystyle \)	\(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)\)	\(=\)	\(\displaystyle \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a c - b d}\right)^2 + \left({a d + b c}\right)^2\)	\(\displaystyle \)

This is an example of a more general identity:

	\(\displaystyle \)	\(\displaystyle \left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right)\)	\(=\)	\(\displaystyle \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a c - n b d}\right)^2 + n \left({a d + b c}\right)^2\)	\(\displaystyle \)

Corollary

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be integers.

Then:

$\displaystyle \prod_{j=1}^n \left({a_j^2 + b_j^2}\right) = c^2 + d^2$

where $c, d \in \Z$.

What this says is that the product of any number of sums of two squares is also a sum of two squares.

More generally:

$\displaystyle \prod_{j=1}^n \left({a_j^2 + n b_j^2}\right) = c^2 + n d^2$

where $c, d \in \Z$.

That is, the set of all numbers of the form $x^2 + n y^2$ is closed under multiplication.

Proof

	\(\displaystyle \)	\(\displaystyle \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2\)	\(=\)	\(\displaystyle a^2 c^2 + 2nabcd + n^2 b^2 d^2 + n a^2 d^2 - 2nabcd + n b^2 c^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^2 c^2 + n a^2 d^2 + n b^2 c^2 + n^2 b^2 d^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right)\)	\(\displaystyle \)

$\blacksquare$

Setting $b = -b$ in the above gives the identity:

$\left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right) = \left({a c - n b d}\right)^2 + n \left({a d + b c}\right)^2$

The identities:

	\(\displaystyle \)	\(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)\)	\(=\)	\(\displaystyle \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a c - b d}\right)^2 + \left({a d + b c}\right)^2\)	\(\displaystyle \)

follow from the above by setting $n = 1$.

$\blacksquare$

Proof of Corollary

Follows by induction from the main result.

$\blacksquare$

Note

This identity is also known as Fibonacci's Identity, and is a special case for $n = 2$ of Lagrange's Identity.

Source of Name

This entry was named for Brahmagupta‎ and Leonardo Fibonacci‎.

Both of these described this identity in their writings:

• 628: Brahmagupta: Brahmasphutasiddhanta (The Opening of the Universe)
• 1225: Fibonacci: Liber quadratorum (The Book of Squares)