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Wednesday, November 23, 2016

Mathematical Analysis - Exercise 1.3

Problem:


Solution:

Suppose $ n = ab $ is not prime, then we can write $ 2^{ab} = (2^{a})^b $. Applying the identity we just proved in the previous problem, we have:

$ 2^{ab} - 1 =  (2^{a})^b - 1 =  (2^{a})^b - 1^b = (2^a - 1)(\cdots) $.

Therefore if $ a \ne 1 $, we have a non-trivial factor for $ 2^{ab} - 1 $.

The contradiction show $ n $ has to be prime.

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