Prove by induction that $\frac{x^{n}-1}{x-1}$ is an integer

Question

For all natural numbers, $x>1$ and $n$, $x^{n}-1$ is divisible by ${x-1}$

Tried to get all the possible equalities and stuff but i'm stucked on it. How to proceed?

Answer 1

I'll assume you mean $x^n-1$ is divisible by $x-1$. With that, the induction step is: $$ x^{n+1}-1=x^{n+1}-x^n+x^n-1=x^n(x-1)+x^n-1. $$ Consider the rightmost expression. The first term there is clearly divisible by $x-1$ whereas the second term is divisible by $x-1$ because of the induction hypothesis. It remains to check the base case and write up a formal solution.

p.s. Induction is actually unneeded if you know the formula $$ x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+1). $$

Answer 2

Note that $x \equiv_{x-1} 1$. So then $$x^n - 1 \equiv_{x-1} 1^n- 1 \equiv_{x-1} 1 - 1 \equiv_{x-1} 0$$ Then $x^n - 1 \equiv_{x-1} 0 \implies (x - 1) \mid (x^n - 1)$.

Here's another proof: $$x^n - 1 = \sum_{i=0}^{n-1} \left(x^{n-i} - x^{n-i-1} \right)= \sum_{i=0}^{n-1} \left((x-1)\left(x^{n-1-i}\right)\right) = (x-1)\sum_{i=0}^{n-1} \left(x^{n-1-i}\right)$$

Answer 3

Also, $\frac{x^n-1}{x-1}=x^{n-1}+\cdots+x+1$. I suppose you could prove that by induction.