Calculate:
$latex (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{32} +1)
$

Let
$latex x = (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{2n} +1)
$

Or:
$latex displaystyle
x = sum_{n=0}^{n}(3^{2n} + 1) $

Quite messy to expand out:

$latex displaystyle {
sum_{n=0}^{n} (3^{2n})
+
sum_{n=0}^{n}(1)
= ….
}
$

This 14-year-old vienamese student in Berlin – Hyyen Nguyen Thi Minh discovered a smart trick using the identity:
$latex displaystyle { (a -1).(a + 1) = a^{2} – 1}$
or more general,
$latex displaystyle boxed {
(a^{n} -1).(a^{n} + 1) = a^{2n} – 1
}$

He multiplies x by (3-1):

$latex
x. (3-1) = (3-1)(3+1). (3^{2} +1)… (3^{2n} + 1)
$
$latex 2x = (3^{2} -1). (3^{2} +1)…(3^{2n} + 1)
$

$latex 2x = (3^{4} -1).(3^{4} +1) … (3^{2n} + 1)
$
.
.
.

$latex 2x = (3^{4n} -1) $

$latex displaystyle boxed
{
x = (3^{4n}…