Question #2114

Use induction to prove that x[sup](2n-1)[/sup]+y[sup](2n-1)[/sup] is divisible by x+y for n is the element of natural numbers

Expert's answer

For n =1:

<img src="/cgi-bin/mimetex.cgi?x%5E%7B2n-1%7D+y%5E%7B2n-1%7D%20=%20x+y" title="x^{2n-1}+y^{2n-1} = x+y">& - the condition is satisfied.

Let's assume that <img src="http://latex.codecogs.com/gif.latex?x%5E%7B2n-1%7D+y%5E%7B2n-1%7D" title="x^{2n-1}+y^{2n-1}">

is divisible by (x+y), let's prove it for n+1:

<img src="http://latex.codecogs.com/gif.latex?x%5E%7B2%28n+1%29-1%7D+y%5E%7B2%28n+1%29-1%7D%20=%20x%5E2x%5E%7B2n-1%7D+y%5E2y%5E%7B2n-1%7D%20=%20x%5E2x%5E%7B2n-1%7D+%28y%5E2%20+%20x%5E2%20-%20x%5E2%29y%5E%7B2n-1%7D=%5C%5C%20=x%5E2%28x%5E%7B2n-1%7D+y%5E%7B2n-1%7D%29%20+%28x+y%29%28x-y%29y%5E%7B2n-1%7D" title="x^{2(n+1)-1}+y^{2(n+1)-1} = x^2x^{2n-1}+y^2y^{2n-1} = x^2x^{2n-1}+(y^2 + x^2 - x^2)y^{2n-1}=\\ =x^2(x^{2n-1}+y^{2n-1}) +(x+y)(x-y)y^{2n-1}">

The first term is divisible by assumption and it's shown that the second term is divisible by (x+y) too.

<img src="/cgi-bin/mimetex.cgi?x%5E%7B2n-1%7D+y%5E%7B2n-1%7D%20=%20x+y" title="x^{2n-1}+y^{2n-1} = x+y">& - the condition is satisfied.

Let's assume that <img src="http://latex.codecogs.com/gif.latex?x%5E%7B2n-1%7D+y%5E%7B2n-1%7D" title="x^{2n-1}+y^{2n-1}">

is divisible by (x+y), let's prove it for n+1:

<img src="http://latex.codecogs.com/gif.latex?x%5E%7B2%28n+1%29-1%7D+y%5E%7B2%28n+1%29-1%7D%20=%20x%5E2x%5E%7B2n-1%7D+y%5E2y%5E%7B2n-1%7D%20=%20x%5E2x%5E%7B2n-1%7D+%28y%5E2%20+%20x%5E2%20-%20x%5E2%29y%5E%7B2n-1%7D=%5C%5C%20=x%5E2%28x%5E%7B2n-1%7D+y%5E%7B2n-1%7D%29%20+%28x+y%29%28x-y%29y%5E%7B2n-1%7D" title="x^{2(n+1)-1}+y^{2(n+1)-1} = x^2x^{2n-1}+y^2y^{2n-1} = x^2x^{2n-1}+(y^2 + x^2 - x^2)y^{2n-1}=\\ =x^2(x^{2n-1}+y^{2n-1}) +(x+y)(x-y)y^{2n-1}">

The first term is divisible by assumption and it's shown that the second term is divisible by (x+y) too.

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