Brahmagupta-Fibonacci Identity/Proof 2

Theorem

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

Then:

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

Proof

From the more general version of Brahmagupta-Fibonacci Identity:

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

The result follows by setting $n = 1$.

$\blacksquare$

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)

However, it appeared earlier than either of those in Diophantus of Alexandria's Arithmetica of the third century C.E.