Brahmagupta-Fibonacci Identity/Extension
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 |
$\blacksquare$