Brahmagupta-Fibonacci Identity/General/Corollary
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Corollary to General Version of Brahmagupta-Fibonacci Identity
Let $a, b, c, d, n$ be numbers.
Then:
- $\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c - n b d}^2 + n \paren {a d + b c}^2$
Proof
\(\ds \paren {a^2 + n b^2} \paren {c^2 + n d^2}\) | \(=\) | \(\ds \paren {a c + n b d}^2 + n \paren {a d - b c}^2\) | General Version of Brahmagupta-Fibonacci Identity | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a^2 + n \paren {-b}^2} \paren {c^2 + n d^2}\) | \(=\) | \(\ds \paren {a c + n \paren {-b} d}^2 + n \paren {a d - \paren {-b} c}^2\) | substituting $-b$ for $b$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a^2 + n b^2} \paren {c^2 + n d^2}\) | \(=\) | \(\ds \paren {a c - n b d}^2 + n \paren {a d + b c}^2\) |
$\blacksquare$
Source of Name
This entry was named for Brahmagupta and Leonardo Fibonacci.