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Prove using induction that for all non-negative integers n and for all integers $ x > 1 $, $ x^n - 1 $ is divisible by $ x - 1 $.

Step 1: We will prove this using induction on n. Step 2: Assume the claim is true when $ n = 1 $. $$ x^{n+1} - 1 = x(x^n - 1) + (x - 1) $$

Can someone help me with this further?

Brian M. Scott
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user3434743
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    http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b – lab bhattacharjee Oct 21 '14 at 15:26
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    The sum of two terms that are divisible by $x-1$ is also divisible by $x-1$. – Henno Brandsma Oct 21 '14 at 15:27
  • If $x^n-1$ is divisible by $x-1$ you can get common factor. – mfl Oct 21 '14 at 15:27
  • Your step 2 should say "Assume the claim is true for some positive integer n; we'll show it's true for $n + 1$." And before it should be Step 1.5: "We'll first show the claim is true for $n = 1$", with an explanation. To continue the general case proof, what can you say about the $x^n -1$ on the right hand side? What degree is it? – John Hughes Oct 21 '14 at 15:27
  • you've done it, since $x-1$ divides $x-1$, obviously, and $x-1$ divides $x^n-1$ by the inductive hypothesis. – David Holden Oct 21 '14 at 15:28
  • You don't assume it's true for $n=1$, you show it. So what do you get for $n=1$? – Henno Brandsma Oct 21 '14 at 15:28

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

Alternative: Actually We don't need to use Induction,

Let $f(x)=x^n-1$ Now $f(1)=0$, That means $1$ is a root if $f(x)$ so $f(x)=(x-1)\times(?)$ ( By '$?$' I mean some another factors as $f(x)$ is polynomial of degree $n$ )