Proof of the Brahmagupta–Fibonacci identity

Proof

We expand both sides, using the laws of a commutative ring.

The left side is $(a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$ .
We have $(ac - bd)^2 = a^2c^2 - 2abcd + b^2d^2$ and $(ad + bc)^2 = a^2d^2 + 2abcd + b^2c^2$ .
Adding the two lines of step 2, the terms $\pm 2abcd$ cancel. The sum is $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2$ , which is the left side by step 1.
Next, $(ac + bd)^2 = a^2c^2 + 2abcd + b^2d^2$ and $(ad - bc)^2 = a^2d^2 - 2abcd + b^2c^2$ .
Adding the two lines of step 4, the terms $\pm 2abcd$ cancel again. The sum is again the left side.
Steps 3 and 5 prove both equalities.

Idea

Each square on the right gives two of the four products $a^2c^2, a^2d^2, b^2c^2, b^2d^2$ and a cross term $\pm 2abcd$ . The signs are chosen so that the cross terms cancel. Over the complex numbers this is $\abs{z}^2 \abs{w}^2 = \abs{zw}^2$ for $z = a + bi$ and $w = c + di$ .

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Group concept
Ring definition
Commutative ring definition
Proof of the Brahmagupta–Fibonacci identity proof

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Tag	`8VC5HCNAJF`
URI	`m2://8VC5HCNAJF`
Kind	proof
Writing	W0 stub
MSC	11E25
Proves	Brahmagupta–Fibonacci identity
Uses	Commutative ring
Changed	4 November 2026

Proof of the Brahmagupta–Fibonacci identity

Proof

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Where this leads

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