Prove by mathematical induction: $x^n - y^n$ is divisible by $x - y$, for all positive integers $n$.

Question

Here is what I have so far:

Base: When $n=1$, we have $x^1 - y^1 = x - y$. Hence, P(1) is true.

Inductive hypothesis: We assume that P(k) is true: $x^k - y^k$ is divisible by $x - y$. That is: $x^k-y^k=(x - y)z$, for some integer $z$

Inductive step: We show that P(k+1) is true. That is:

$x^{k+1} - y^{k+1} = (x - y)z$

I don't quite know where to go from there.

Does this answer your question? Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ — Alejandro Bergasa Alonso, Feb 01 '21 at 15:07

score 7 · Answer 1 · answered Feb 01 '21 at 14:36

7

The induction hypothesis should be “$x-y$ divides $x^k-y^k$”, not that they're equal.

Hint: $x^{k+1}-y^{k+1}=x^{k+1}-xy^{k}+xy^k-y^{k+1}$

Alternate hint: from the induction hypothesis you have $x^k-y^k=(x-y)z$, so $y^k=x^k-(x-y)z$ and therefore $$ x^{k+1}-y^{k+1}=x^{k+1}-x^ky+(x-y)yz $$

answered Feb 01 '21 at 14:36

egreg

1

Sorry if this is a dumb question but where did z on the RHS come from? – Andy Feb 01 '21 at 14:51
$z$ is a polynomial defined by $z=\dfrac{x^n-y^n}{x-y}$. – ultralegend5385 Feb 01 '21 at 15:07
@Andy $z$ exists by the induction hypothesis that asserts divisibility. You need not know what it looks like, just that it exists. – egreg Feb 01 '21 at 15:10

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