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$(1)$ Let a and b be positive integers and suppose b is odd. Show that $1 + a^b$ is divisible by $a+1$. $\;\quad( $Suggested method is using the geometric sum formula.)

$(2)$ Let k be a positive integer. Show that if $2^k + 1$ is prime, then $k=2^n$ for some $n \in \mathbb N$.

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  • $\begingroup$ For the first one: let $m=a+1$. Now, you just need to prove that $1+(m-1)^b$ is a multiple of $m$. Use the binomial theorem. $\endgroup$ – Akiva Weinberger Nov 12 '14 at 15:34
  • $\begingroup$ Typically, try to limit your questions to one per post. Arguably, these might be seen as sufficiently related, so I'm not going to make an issue of it. $\endgroup$ – Namaste Nov 12 '14 at 15:39
  • $\begingroup$ Ok sorry. Will do that in the future :) $\endgroup$ – mathgrad93 Nov 12 '14 at 19:18
  • $\begingroup$ See $a^n - 1 \mid a^m - 1$ if and only if $n \mid m$ for a more general result. $\endgroup$ – punctured dusk Apr 15 '15 at 8:45
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If $k$ is not a power of two, so $k$ has an odd factor $b$, write $2^k+1=a^b+1$ for suitable $a$.
How does that answer question 2?

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For your first question:

$ \sum_{k=0}^{b-1} (-a)^k = \frac{1-(-a)^b}{1+a} = \frac{1+a^b}{1+a} $ , since b is odd: $-(-a)^b = -(-1)^b*a^b = 1*a^b =a^b $

=> $(1+a)*[\sum_{k=0}^{b-1} (-a)^k] = 1 + a^b $

Hence $(a+1)|(a^b +1)$

Hint: For the second question, you might want to use the first question..

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