# Not a power of 2

• November 18th 2009, 11:32 AM
james_bond
Not a power of 2
Show that $k^n+k^{n-1}+\ldots +k+1$ can't be a power of 2 if $k>1$ ( $n,k\in\mathbb{Z^+}$).
• November 18th 2009, 03:29 PM
tonio
Quote:

Originally Posted by james_bond
Show that $k^n+k^{n-1}+\ldots +k+1$ can't be a power of 2 if $k>1$ ( $n,k\in\mathbb{Z^+}$).

This is false for n = 1 and k = 3 or 7, or 15 or 31 or....

Tonio
• November 19th 2009, 06:57 PM
NonCommAlg
the claim would be true if we also assume that n > 1. the proof is quite easy: let $f_n(k)=k^n + k^{n-1} + \cdots + k + 1, \ \ k > 1, \ n>1.$ it's clear that if $n$ is even, then $f_n(k)$ is odd.

so we may assume that $n$ is odd. the proof now is by induction over $n.$ we have $f_3(k)=k^3+k^2+k+1=(k+1)(k^2+1).$ suppose $f_3(k)$ is a power of 2. then $k=2r+1,$ for

some $r \geq 1,$ and so $k^2+1=4r(r+1)+2,$ which is never a power of 2 because $r > 0.$ if $n=2m+1 > 3,$ then the identity $f_{2m+1}(k)=f_m(k) (k^{m+1} + 1)$ completes the proof.
• November 19th 2009, 06:59 PM
Drexel28
Quote:

Originally Posted by NonCommAlg
the claim would be true if we also assume that n > 1. the proof is quite easy: let $f_n(k)=k^n + k^{n-1} + \cdots + k + 1, \ \ k > 1, \ n>1.$ it's clear that if $n$ is even, then $f_n(k)$ is odd.

so we may assume that $n$ is odd. the proof now is by induction over $n.$ we have $f_3(k)=k^3+k^2+k+1=(k+1)(k^2+1).$ suppose $f_3(k)$ is a power of 2. then $k=2r+1,$ for

some $r \geq 1,$ and so $k^2+1=4r(r+1)+2,$ which is never a power of 2 because $r > 0.$ if $n=2m+1 > 3,$ then the identity $f_{2m+1}(k)=f_m(k) (k^{m+1} + 1)$ completes the proof.

Not to sound like a creeper, but I notice how long it takes you to write yoru responses to post. You take very long to write up your proofs. Is that because you are being careful?
• November 19th 2009, 11:22 PM
NonCommAlg
Quote:

Originally Posted by Drexel28
Not to sound like a creeper, but I notice how long it takes you to write yoru responses to post. You take very long to write up your proofs. Is that because you are being careful?

ok, i guess i misunderstood your question. i get it now. yes, i like my response to be as accurate as possible. also sometimes while i'm replying to a question i find a better solution and so i start

my response all over again.
• November 19th 2009, 11:24 PM
Drexel28
Quote:

Originally Posted by NonCommAlg
ok, i guess i misunderstood your question. i get it now. yes, i like my response to be as accurate as possible. also sometimes while i'm replying to a question i find a better solution and so i start

my response all over again.

Haha, it's cool NCA. I just always wondered why it took you so long. There were a couple times you replied to some of my threads and it was all I could take to wait. (I always think you are going to go "This is bull____!")
• November 24th 2009, 11:11 AM
james_bond
Quote:

Originally Posted by tonio
This is false for n = 1 and k = 3 or 7, or 15 or 31 or....

Tonio

Yeah sorry I meant $n>1$...