Question #b809d

Question

815 views around the world

You can reuse this answer
Creative Commons License

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.