Brahmagupta-Fibonacci Identity/Extension/Proof 3

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Extension to Brahmagupta-Fibonacci Identity

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

Then:

$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$

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

Let $z_j = a_j + i b_j$ for each $j = 1, 2, \ldots, n$.

Let $\ds c + i d = \prod_{j \mathop = 1}^n z_j$.

Then:

\(\ds c + i d\) \(=\) \(\ds \prod_{j \mathop = 1}^n z_j\)
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 1}^n \paren {a_j + i b_j}\)


As $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are integers, so are $c$ and $d$.


Thus:

\(\ds c^2 + d^2\) \(=\) \(\ds \cmod {c + i d}^2\) Definition of Complex Modulus
\(\ds \) \(=\) \(\ds \cmod {\prod_{j \mathop = 1}^n z_j}^2\) $\ds c + i d = \prod_{j \mathop = 1}^n z_j$
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 1}^n \cmod {z_j}^2\) Complex Modulus of Product of Complex Numbers: General Result
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 1}^n \cmod {a_j + i b_j}^2\) $z_j = a_j + i b_j$
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2}\) Definition of Complex Modulus

$\blacksquare$


Sources

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