Thread: Not a power of 2

1. Not a power of 2

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

2. 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

3. 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.

4. 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?

5. 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.

6. 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____!")

7. 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 $n>1$...