# For positive integers a, b, with b odd, show that $(a+1)\mid (1 + a^b)$.

$(1)$ 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 $n \in \mathbb N$.

• 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. – Akiva Weinberger Nov 12 '14 at 15:34
• 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. – Namaste Nov 12 '14 at 15:39
• Ok sorry. Will do that in the future :) – mathgrad93 Nov 12 '14 at 19:18
• See $a^n - 1 \mid a^m - 1$ if and only if $n \mid m$ for a more general result. – punctured dusk Apr 15 '15 at 8:45

If $k$ is not a power of two, so $k$ has an odd factor $b$, write $2^k+1=a^b+1$ for suitable $a$.
$\sum_{k=0}^{b-1} (-a)^k = \frac{1-(-a)^b}{1+a} = \frac{1+a^b}{1+a}$ , 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 $(a+1)|(a^b +1)$