Brahmagupta-Fibonacci Identity

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Theorem

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

Then:

$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$


Corollary

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


General result

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

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


Extension

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

Then:

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

where $c, d \in \Z$.

That is: the product of any number of sums of two squares is also a sum of two squares.


Proof 1

\(\displaystyle \) \(\) \(\displaystyle \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({a c}\right)^2 + 2 \left({a c}\right) \left({b d}\right) + \left({b d}\right)^2}\right) + \left({\left({a d}\right)^2 - 2 \left({a b}\right) \left({c d}\right) + \left({b c}\right)^2}\right)\) $\quad$ Square of Sum, Square of Difference $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a^2 c^2 + 2 a b c d + b^2 d^2 + a^2 d^2 - 2 a b c d + b^2 c^2\) $\quad$ multiplying out $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)\) $\quad$ $\quad$

$\blacksquare$


Proof 2

From the more general version of Brahmagupta-Fibonacci 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 result follows by setting $n = 1$.

$\blacksquare$


Proof 3

Lagrange's Identity gives:

$\displaystyle \left({\sum_{k \mathop = 1}^n {a_k}^2}\right) \left({\sum_{k \mathop = 1}^n {b_k}^2}\right) - \left({\sum_{k \mathop = 1}^n a_k b_k}\right)^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \left({a_i b_j - a_j b_i}\right)^2$


Setting $n = 2$:

\(\displaystyle \left({\sum_{k \mathop = 1}^2 {a_k}^2}\right) \left({\sum_{k \mathop = 1}^2 {b_k}^2}\right) - \left({\sum_{k \mathop = 1}^2 a_k b_k}\right)^2\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^1 \sum_{j \mathop = i + 1}^2 \left({a_i b_j - a_j b_i}\right)^2\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({ {a_1}^2 + {a_2}^2}\right) \left({ {b_1}^2 + {b_2}^2}\right) - \left({a_1 b_1 + a_2 b_2}\right)^2\) \(=\) \(\displaystyle \left({a_1 b_2 - a_2 b_1}\right)^2\) $\quad$ expanding out summations $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right) - \left({a c + b d}\right)^2\) \(=\) \(\displaystyle \left({a d - b c}\right)^2\) $\quad$ renaming $a_1 \to a, a_2 \to b, b_1 \to c, b_2 \to d$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)\) \(=\) \(\displaystyle \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2\) $\quad$ $\quad$

$\blacksquare$


Proof 4

Let $z_1 = a + b i, z_2 = c + d i$ be complex numbers.

Let $\left\lvert{z}\right\rvert$ denote the complex modulus of a given complex number $z \in \C$.


By definition of complex multiplication:

$(1): \quad z_1 z_2 = \left({a c - b d}\right) + \left({a d + b c}\right) i$


Then we have:

\(\displaystyle \left\lvert{z_1 z_2}\right\rvert\) \(=\) \(\displaystyle \left\lvert{z_1}\right\rvert \cdot \left\lvert{z_2}\right\rvert\) $\quad$ Complex Modulus of Product of Complex Numbers $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left\lvert{\left({a + b i}\right) \left({c + d i}\right)}\right\rvert^2\) \(=\) \(\displaystyle \left\lvert{a + b i}\right\rvert^2 \cdot \left\lvert{c + d i}\right\rvert^2\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({a c - b d}\right)^2 + \left({a d + b c}\right)^2\) \(=\) \(\displaystyle \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)\) $\quad$ Definition of Complex Modulus, and from $(1)$ $\quad$

$\blacksquare$


Also known as

This identity is also known as Fibonacci's Identity.


Also see

Lagrange's Identity, of which this the case where $n = 2$.


Source of Name

This entry was named for Brahmagupta‎ and Leonardo Fibonacci‎.


Historical Note

Both Brahmagupta‎ and Leonardo Fibonacci‎ described what is now known as the Brahmagupta-Fibonacci Identity in their writings:

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


Sources