Wednesday, 6 March 2013

WEDNESDAY, 6 MARCH 2013

Today is the $65^{th}$ day of the year.

$65 = 5 \times 13$

The Brahmagupta or Fibonacci Identity states that
$$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$$
and
$$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$$
Given that we know that
$65 = 5 \times 13$
$5 = 1 + 4 = 1^2 + 2^2$
$13 = 4 + 9 = 2^2 + 3^2$
$65 = (1^2 + 2^2)(2^2 + 3^2)$ giving $a = 1$, $b = 2$, $c = 2$ and $d = 3$ in the left hand side of either form of the identity.

We can then conclude from the first identity that
$65 = (1 \times 2 + 2 \times 3)^2 + (1 \times 3 - 2 \times 2)^2$
$65 = (2 + 6)^2 + (3 - 4)^2$
$65 = 8^2 + (-1)^2$
$65 = 64 + 1$

Using the second identity we can conclude that
$65 = (1 \times 2 - 2 \times 3)^2 + (1 \times 3 + 2 \times 2)^2$
$65 = (2 - 6)^2 + (3 + 4)^2$
$65 = (-4)^2 + 7^2$
$65 = 16 + 49$

Leading us to the conclusion that
$65 = 1^2 + 8^2 = 4^2 + 7^2$