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Old 2020-10-14, 17:20   #10
bur
 
Aug 2020

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Having looked at the proof, what I don't get is, why is 2^{(2r+1)s}=(2^s + 1)(...)?

Wolfram expressed it differently as (2^a)^b = (2^a + 1)(2^{a(b-1)}-2^{a(b-2)}+...). For that proof, I don't understand why 2^n = (2^a)^b if n = a*b with b being an odd integer. That it's equivalent to the product, I also don't get.

Last fiddled with by bur on 2020-10-14 at 17:24
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