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My teacher once said, for any positive number $\ n, $ $\ n^k - 1 $ would always have $\ n-1 $ as a factor for all positive odd values of $\ k $. Could anyone tell me the proof? I have written my approach below.

Assuming $n-1$ to be a factor.

$n^k - 1 = (n-1)x $

$n^k = (n-1)x + 1 $

$k = \lg( (n-1)x + 1 ) / \lg (n) $

If my approach is right could you tell me where should I go from here?

Edit : This clears my doubt.

Edit 2 : My question is "why for any positive number $\ n, $ $\ n^k + 1 $ would always have $\ n+1 $ as a factor for all positive odd values of $\ k $ ".

Community
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    See http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b/188710 – lab bhattacharjee Jun 23 '16 at 08:20
  • Concerning the title: $k$ does not have to be odd for this (as you probably will have noticed by now). Maybe you mixed it up with $n+1$ being a factor of $n^k+1$. That will be the case if $k$ is odd. – drhab Jun 23 '16 at 08:26
  • yes :) @drhab I get it now. –  Jun 23 '16 at 08:28
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    If $k$ is odd then $n^k+1=n^k-(-1)^k$ so that $n+1=n-(-1)$ will be a factor of it. – drhab Jun 23 '16 at 08:35
  • I didnt get it, could you clear it further? @drhab the $ n^k + 1 $ case. –  Jun 23 '16 at 08:42
  • In the link you find an explanation for $a-b$ being a factor of $a^k-b^k$. Now substitute $a=n$ and $b=-1$. Apparantly $n-(-1)$ is a factor of $n^k-(-1)^k$. Here $n+1=n-(-1)$ and if moreover $k$ is odd then $(-1)^k=-1$ so that $n^k-(-1)^k=n^k+1$. – drhab Jun 23 '16 at 08:50
  • @sidgupta234 Use factor theorem for the polynomials $f(n)=n^k -1$,(if $k \in \mathbb{N}$) when $n=1$ and $g(n)=n^k+1$,(if $k$ is odd) when $n=-1$ and see what you get. – Roby5 Jun 23 '16 at 08:53
  • @sidgupta234 : for your second edit, you should ask a new question. – Watson Jun 23 '16 at 11:43

4 Answers4

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The identity \begin{align} n^k-1&=(n-1)(n^{k-1}+\dots+n+1)\tag{1} \end{align} can be proved with an easy induction on $k$ ($k$ doesn't have to be odd):

This is a tautology for $k=1$.

Suppose, for the inductive step, we have $(1)$ for some $k$. Then \begin{align*} n^{k+1}-1&=(n^{k+1}-n^k)+(n^k-1)=n^k(n-1)+(n-1)(n^{k-1}+\dots+n+1)\\ &=(n-1)(n^k+n^{k-1}+\dots+n+1). \end{align*}

Bernard
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$n^k-1=(n-1)(n^{k-1}+n^{k-2}+n^{k-3}+\dots+n+1)$ is true for all positive integers $k$

But I suspect what you remember is actually

$n^k+1=(n+1)(n^{k-1}-n^{k-2}+n^{k-3}-\dots-n+1)$ is true for all odd positive integers $k$.

The reason the second statement cannot be made for even $k$ is that the alternating signs would lead to a final $-1$, which when multiplied by $+1$ would give $-1$ rather than the $+1$ on the left hand side. So you also have

$n^k-1=(n+1)(n^{k-1}-n^{k-2}+n^{k-3}-\dots+n-1)$ is true for all even positive integers $k$

and

$n^k-1=(n+1)(n-1)(n^{k-2}+n^{k-4}+\dots+1)$ is true for all even positive integers $k$

Henry
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First of all this is true for any positive integer $k$, I have no idea why he said odd $k$.

There are many ways to prove it. The most straight forward is to use the fact that:

$$(n-1)(n^{k-1}+\cdots+1) = n^k-1$$

you can also use Euclid algorithm for polynomial division and see that because $n^k-1$ has a zero for $n=1$ you're bound to have $n-1$ as a factor.

A third way is to use induction. Use that

$n^{k+1}-1 = (n^k-1)(n+1) + n - n^k = (n^k-1)(n+1) - (n^{k-1}-1)n$

to see that if it's true for the exponent $k$ and $k-1$ it's also true for the exponent $k+1$. To finish the induction proof you would have to prove it for $k=1$ (trivial) and $k=2$ (which follows from $n^2-1 = (n-1)(n+1)$ so $n-1$ is a factor).

Your approach on the other hand is probably a dead end for two reasons. First of all you assume that $n-1$ is a factor which isn't very helpful if you're to prove that $n-1$ is a factor. Second you seem to try to solve for $k$ (which isn't useful as $k$ can be any positive integer).

One reason to exclude even $k$ may be that if it's even, say $k=2j$ you could rewrite $n^k-1 = n^{2j}-1 = (n^j-1)(n^j+1)$. But on the other hand if $j$ is odd $(n-1)$ would be a factor of $n^j-1$, otherwise you can repeat the same procedure and eventually you will factor out $n^p-1$ where $p$ is odd.

skyking
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More generally, if $a,b\in\mathbb Z$, $n\in\mathbb Z^+$, then $a-b\mid a^n-b^n$.

It follows from $$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$$

Edit: Since it seems you wanted to ask about the fact that for $n\in\mathbb Z$ with odd $k\in\mathbb Z^+$ we have $n+1\mid n^k+1$:

More generally, for $a,b\in\mathbb Z$ and odd $n\in\mathbb Z^+$ we have $$a^n+b^n=(a+b)\times$$

$$\times \left(a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots-ab^{n-2}+b^{n-1}\right)$$

user236182
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