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Proposition: $x^n-a^n$ is divisible by $x-a$ for all n. Potential proof by induction (which I would like to know whether it is correct):

Suppose n=1. Then $x^n-a^n=x-a$, and $x-a$ is clearly divisible by $x-a$. So the proposition holds for n=1, the base case.

Now suppose the proposition holds for n=k. This means that $x^k-a^k$ is divisible by $x-a$.

Now we have to prove that the proposition holds for n=k+1. Then it has to be proved that $x^{k+1}-a^{k+1}$ is divisible by x-a. $x^{k+1}-a^{k+1}$ is the same as $x^k \cdot x - a^k \cdot a$, which is equivalent to $(x^k-a^k) \cdot ax$. However, it was already established that the proposition holds for n=k, so $x^k-a^k$ is divisible by x-a. This renders the expression $(x^k-a^k) \cdot ax$ divisible by x-a; when this expression is divided by x-a, only $ax$ is left.

I don't know if my proof is correct since I just learned induction.

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  • $\begingroup$ Does this answer your question? $a-b$ divides $a^n-b^n$ . Found using Approach0. Note another duplicate is $x-y$ divides $x^n - y^n$ -- prove by mathematical induction. $\endgroup$ May 7 at 17:12
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    $\begingroup$ You may be interested to know that for a polynomial $p(x)$ it is always true that $p(x)-p(a)$ is divisible by $x-a$. If follows from the statement you are trying to prove, but it can also be shown more briefly using polynomial division. $p(x)-p(a)=q(x)(x-a)+b$ for some polynomial $q(x)$ and some constant remainder $b$. Now set $x=a$ to obtain $b=0$. This conceals the induction (the division algorithm requires it), but is a neat proof and embodies a potentially useful technique. $\endgroup$ May 7 at 17:15

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The step $x^k\cdot x - a^k \cdot a = (x^k - a^k)\cdot ax$ is wrong.

A simple counterexample is $3^2\cdot 3 - 4^2\cdot 4 \neq (3^2 - 4^2)\cdot (3\times4)$.

To correct the proof, substitute $x^k = p + a^k$ with the assumption that $p$ is divisible by $(x-a)$.

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The structure of your proof by induction is correct, however you have an invalid algebraic step. You claim $x^kx - a^ ka = (x^k -a^k)ax$, but that is not true. You will need to change your algebraic steps in the proof, but you would have been right except for that part, so you understand induction alright!

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