Smart Algebraic Technique

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

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

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

Quite messy to expand out:

\displaystyle { \sum_{n=0}^{n} (3^{2n}) +  \sum_{n=0}^{n}(1) = .... }

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

He multiplies x by (3-1):

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

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

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

\displaystyle \boxed  { x =  (3^{4n} -1) / 2 }
When n = 16,
\displaystyle  { x =  (3^{64} -1) / 2  }

x =  1,716,841,910,146,256,242,328,924,544,640

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