Polynomial ,divides and Induction Proof?

Question

$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n } \quad x-y \;\text{ divides }\; x^n-y^n \text{ prove by induction.}$

What I am confused about in this example is when one makes $n=k$. Which variable is made equal to k.Where is $n$. Is it fair to assume that what the actual problem wants you to infer this. $x^n +y^n = x^n+y^n$

(i) Basis Step: $x^n-y^n = x^n-y^n$

$x-y = x-y$

$0=0$

(ii) Inductive step : $x^k-y^k = x^k-y^k$

WNTS: RHS: $n =k+1$

$x^{k+1} + y^{k+1}$

LHS: $x^{k+1} + y^{k+1} +x^k-y^k $

Is this a fair assumption to make I feel as if their is something fundamentally missing from my logic. Any suggestions would be good.

Answer 1

Let say ${ x }^{ n }-{ y }^{ n }$ is dividing by $x-y$ then we have $${ x }^{ n }-{ y }^{ n }=\left( x-y \right) P\left( x \right) $$,where $P(x)$ is some polynomial.Now we should prove that ${ x }^{ n+1 }-{ y }^{ n+1 }$ is also dividing by $x-y$ $${ x }^{ n+1 }-{ y }^{ n+1 }={ x }^{ n+1 }-{ x }^{ n }y+{ x }^{ n }y-{ y }^{ n+1 }=\\{ x }^{ n }\left( x-y \right) +y\underbrace { \left( { x }^{ n }-{ y }^{ n } \right) }_{ \left( x-y \right) P\left( x \right) } ={ x }^{ n }\left( x-y \right) +y\left( x-y \right) P\left( x \right) =\\ =\left( x-y \right) \left( { x }^{ n }+yP\left( x \right) \right) $$ so it means

$${ x }^{ n+1 }-{ y }^{ n+1 }=\left( x-y \right) \left( { x }^{ n }+yP\left( x \right) \right) $$

Answer 2

For the inductive step, suppose $x-y$ divides $x^n-y^n$ for some $n$. You have to prove that, under this hypothesis, $x-y$ divides $x^{n+1}-y^{n+1}$.

Hint:

Rewrite it as $$x^{n+1}-y^{n+1}=x(x^n-y^n)+xy^n - y^{n+1}$$ and make partial factorisations.

Some details:

By the induction hypothesis, $x^n-y^n=(x-y)q(x,y)$, so $$x(x^n-y^n)+xy^n - y^{n+1}=(x-y)q(x,y)+ (x-y)y^n=(x-y)\bigl(q(x,y)+y^n\bigr).$$

Actually, the induction step defines a recurrence relation for the quotient of $x^n-y^n$ by $x-y$, which allows to prove the complete factorisation formula.