Question #b809d

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Answer 1 · 2015-05-21T13:07:34

We have :

$(x^(2n+1) - y^(2n+1))/(x-y) =lambda$ , where $lambda in ZZ$

$=> (x^(2n+1) - y^(2n+1))=lambda(x-y)$

Basis :

With $n=0$ , $(x-y)=1(x-y)=lambda(x-y)$

With $n=1$ , $(x^3-y^3)=(x-y)(x^2+xy+y^2)=lambda(x-y)$

Inductive step :

Let's assume $(x^(2n+1) - y^(2n+1))=lambda(x-y)$ is true for $n in NN$ .

Let's demonstrate that it is true for $n+1 in NN$ :

$x^(2(n+1)+1) - y^(2(n+1)+1)=x^(2n+3) - y^(2n+3)$

$x^(2n+3) - y^(2n+3) = x^(2n+3) + x^(2n+1)y^2-x^(2n+1)y^2 - y^(2n+3)$

$= x^(2n+1)(x^2-y^2) + y^2(x^(2n+1)-y^(2n+1))$

$= x^(2n+1)(x+y)(x-y) + y^2lambda(x-y)$

$= (x-y)(x^(2n+1)(x+y) + lambday^2)$

$= lambda_2(x-y)$ , where $lambda_2=(x^(2n+1)(x+y) + lambday^2) in ZZ$ .

QED.