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

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- Nov 18th 2009, 11:32 AMjames_bondNot a power of 2
Show that $\displaystyle k^n+k^{n-1}+\ldots +k+1$ can't be a power of 2 if $\displaystyle k>1$ ($\displaystyle n,k\in\mathbb{Z^+}$).

- Nov 18th 2009, 03:29 PMtonio
- Nov 19th 2009, 06:57 PMNonCommAlg
the claim would be true if we also assume that n > 1. the proof is quite easy: let $\displaystyle f_n(k)=k^n + k^{n-1} + \cdots + k + 1, \ \ k > 1, \ n>1.$ it's clear that if $\displaystyle n$ is even, then $\displaystyle f_n(k)$ is odd.

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

some $\displaystyle r \geq 1,$ and so $\displaystyle k^2+1=4r(r+1)+2,$ which is never a power of 2 because $\displaystyle r > 0.$ if $\displaystyle n=2m+1 > 3,$ then the identity $\displaystyle f_{2m+1}(k)=f_m(k) (k^{m+1} + 1)$ completes the proof. - Nov 19th 2009, 06:59 PMDrexel28
- Nov 19th 2009, 11:22 PMNonCommAlg
- Nov 19th 2009, 11:24 PMDrexel28
- Nov 24th 2009, 11:11 AMjames_bond