Online Course Discussion Forum

How to do MCIII 5.33 Number Theory?

 
 
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Re: How to do MCIII 5.33 Number Theory?
by Dr. Kevin Wang - Tuesday, 2 June 2020, 3:31 PM
 

I don't think you got the second factor right.  Let $a=1989^{2^t}$, then the first factor is $a-1 = 1989^{2^t}-1$, and the second factor is:

$$a^{n-1} + a^{n-2} + \cdots + a + 1 = 1989^{2^t(n-1)} + 1989^{2^t(n-2)} + \cdots + 1989^{2^t} + 1.$$

When $n$ is an odd number, how many factor of $2$ do you think the second factor has?