How to prove $\forall n \in \mathbb{N}, x^n − y^n \text{ is divisible by } x − y$ using induction.

Question

I'm very stuck on this problem, which is an exercise from our course notes. Any clues or hints or a proof write up would be really helpful! Thanks!

@EeveeTrainer tried rewriting $x^n$ as $k\cdot x$ per the definition of divisibility — user2793618, Feb 16 '20 at 04:15
@J.W.Tanner I'm confused as to how to factor (x-y) out of $x^{n-1}-y^{n-1}$ — user2793618, Feb 16 '20 at 04:22
You don't need to factor $x^{n+1}+y^{n+1}$ into $m\left(x^n+y^n\right)$, nor use only one base case. It is easier with J.W.'s method. — Jam, Feb 16 '20 at 04:34

score 1 · Answer 1 · answered Feb 16 '20 at 04:27

1

Base case: it's true for $n=0$ and $n=1$.

Induction step: assume it's true for $n-1$ and $n$;

i.e., $x^{n-1}-y^{n-1}$ and $x^n-y^n$ are divisible by $x-y$.

Then, since $x^{n+1}-y^{n+1}=(x+y)(x^n-y^n)-xy(x^{n-1}-y^{n-1})$,

it follows that $x^{n+1}-y^{n+1}$ is divisible by $x-y$.

answered Feb 16 '20 at 04:27

J. W. Tanner

Would one base case work for this (assuming that we are using the IH of p(n))? Also I'm still confused as to how the factoring for $x-y$ out of $x^n-y^n$ works. – user2793618 Feb 16 '20 at 04:29
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I used two $n$ values for the base case so I could for the induction step; $x-y$ divides $x^n-y^n$ by the induction hypothesis – J. W. Tanner Feb 16 '20 at 04:33
That seems to almost be like strong induction, what if we just had the IH of p(n)? – user2793618 Feb 16 '20 at 04:35
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I posted another answer that you might like more – J. W. Tanner Feb 16 '20 at 04:49

J. W. Tanner · Accepted Answer · 2020-02-16T04:57:47.340

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Here's an answer with one $n$ value for the base case.

Induction step: it's true for $n=1$.

Now assume it's true for $n$; i.e., $x^{n}-y^{n}$ is divisible by $x-y$.

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

it follows that $x^{n+1}-y^{n+1}$ is divisible by $x-y$.

edited Feb 16 '20 at 04:57

answered Feb 16 '20 at 04:47

J. W. Tanner

But for this solution, if we had $n=0$ and the predicate to be defined as $x^{n-1}-y^{n-1}$, wouldn't this be invalid? – user2793618 Feb 16 '20 at 04:52
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I don't think it was a problem, but I edited to avoid confusion – J. W. Tanner Feb 16 '20 at 04:59
How did you get to factor out like that? – user2793618 Feb 16 '20 at 05:00
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That's another question – J. W. Tanner Feb 16 '20 at 05:03
Touche, but I wouldn't find seeing your work for it – user2793618 Feb 16 '20 at 05:04
Compare $(x^{n+1}-y^{n+1})/(x-y) $ and $(x^n-y^n)/(x-y)$ using the answer for the other question – J. W. Tanner Feb 16 '20 at 05:13

2 Answers2