Brahmagupta-Fibonacci Identity/Extension

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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.


General Version

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

Then:

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

for some $c, d \in \Z$.

That is: the set of all integers of the form $a^2 + m b^2$ is closed under multiplication.


Proof 1

From the extension to the general Brahmagupta-Fibonacci Identity:

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

for some $c, d \in \Z$, for all $m \in \Z$.

The result follows by setting $m = 1$.

$\blacksquare$


Proof 2

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

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

for some $c, d \in \Z$.


$\map P 1$ is the trivial case:

\(\ds \prod_{j \mathop = 1}^1 \paren { {a_j}^2 + {b_j}^2}\) \(=\) \(\ds {a_1}^2 + {b_1}^2\)
\(\ds \) \(=\) \(\ds c^2 + d^2\) setting $c = a_1$ and $d = b_1$

Thus $\map P 1$ is seen to hold.


Basis for the Induction

$\map P 2$ is the case:

\(\ds \prod_{j \mathop = 1}^2 \paren { {a_j}^2 + {b_j}^2}\) \(=\) \(\ds \paren { {a_1}^2 + {b_1}^2} \paren { {a_2}^2 + {b_2}^2}\)
\(\ds \) \(=\) \(\ds \paren {a_1 a_2 + b_1 b_2}^2 + \paren {a_1 b_2 - b_1 a_2}^2\) Brahmagupta-Fibonacci Identity
\(\ds \) \(=\) \(\ds c^2 + d^2\) setting $c = a_1 a_2 + b_1 b_2$ and $d = a_1 b_2 - b_1 a_2$

Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

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

for some $c, d \in \Z$.


from which it is to be shown that:

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

for some $c, d \in \Z$.


Induction Step

This is the induction step:


\(\ds \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + {b_j}^2}\) \(=\) \(\ds \prod_{j \mathop = 1}^k \paren { {a_j}^2 + {b_j}^2} \paren { {a_{k + 1} }^2 + {b_{k + 1} }^2}\)
\(\ds \) \(=\) \(\ds \paren { {c'}^2 + {d'}^2} \paren { {a_{k + 1} }^2 + {b_{k + 1} }^2}\) Induction Hypothesis: for some $c', d' \in \Z$
\(\ds \) \(=\) \(\ds c^2 + m d^2\) Basis for the Induction: for some $c, d \in \Z$


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore, for all $n \in \Z_{>0}$:

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

for some $c, d \in \Z$.

$\blacksquare$


Proof 3

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

$\blacksquare$