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Prove that for all natural number $x$ and $n$, $x^n - 1$ is divisible by $x-1$.

So here's my thoughts: it is true for $n=1$, then I want to prove that it is also true for $n-1$

then I use long division, I get:

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

so the left side is divisible by $x-1$ by hypothesis, what about the right side?

Ian Mateus
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Lazy Frog
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    You want to prove it is true for $n+1$, not $n-1$. You are pretty close though: you just proved that if $x^{n-1}-1$ is divisible by $x-1$, then $x^n-1$ also is divisible. This completes the proof. – Ian Mateus Nov 24 '13 at 22:27

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You seem to be trying to prove exactly the opposite of what you should be proving. In effect you’re assuming that $x^n-1$ is divisible by $x-1$ and showing that $x^{n-1}-1$ is also divisible by $x-1$. You should be going in the other direction: for the induction step you want to show that if $x^n-1$ is divisible by $x-1$, then $x^{n+1}-1$ is also divisible by $x-1$. With very minor modification you already have the necessary algebra:

$$x^{n+1}-1=x\left(x^n-1\right)+(x-1)\;.$$

How can you use this to show that if $x^n-1$ is divisible by $x-1$, then so is $x^{n+1}-1$?

Brian M. Scott
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So first you can't assume that the left hand side is divisible by $x-1$ but for the right hand side we have that $x-1$ divides $x-1$ and by the induction hypothesis we have that $x-1$ divides $x^{n-1}-1$ so what can you conclude about the left hand side.

Stone
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  • well, if we assume $x^{n-1} -1$ is divisible by $x-1$, then is $x(x^{n-1}-1)$ also divisible? Why can we just assume it is correct? – Lazy Frog Nov 24 '13 at 22:44
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let $P(n)$ be the assertion that $x^n-1$ is divisible by $x-1$.

we know that $P(0)$ is true. suppose that for some $n$ we have $P(n)$. then there is a number $k$ such that:

$$ x^n-1 = k(x-1) $$ now add $x^{n+1} - x^n = x^n(x-1)$ to both sides of the equation. this will give you the inductive step from $P(n)$ to $P(n+1)$.

David Holden
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  • sorry, the previous answers were not visible when i wrote the above. i will leave it here just in case the slight difference in formulation helps your understanding, but Brian's answer has all you need. you can improve the look of your post by enclosing the mathematical expressions between $ signs, btw – David Holden Nov 24 '13 at 22:29