Show that x^n - y^n is divisible by (x-y)

  • Thread starter Thread starter bodensee9
  • Start date Start date
bodensee9
Messages
166
Reaction score
0
Hello:

Can someone provide hints on the following? I am supposed to show that

x^n - y^n is divisible by (x-y), where x and y are integers and y < x.

I got to where I would need to show that

x^(n+1) - y^(n+1) = some K*(x-y).

But I am not sure what to do next?

By brute force I think the division would need to be in the form of x^n + yx^(n-1) ... + y^(n-1)

but how would I show something like that?

Thanks!
 
Physics news on Phys.org


You seem to have a pretty good idea of what the quotient K should be. For example x^3-y^3=(x-y)*(x^2+xy+y^2). Multiply the right side out, x*(x^2+xy+y^2)-y(x^2+xy+y^2)=x^3+x^2y+xy^2-(yx^2+x^2y+y^3). Notice how the extra terms cancel in pairs? Can you generalize that to n>3, as you said almost correctly K=x^(n-1)+x^(n-2)y+x^(n-3)y^2+...+x*y^(n-2)+y^(n-1).
 


oh got it.

I think the middle term would be like + x*y^n - x*y^n and then I can make the equation into

x*(x^n - y^n) + y^n*(x-y) and since x and y are integers, I am done.

Thanks!
 


What does

<br /> x^{n+1} - y^{n+1} = \left(x^{n+1} - x^n\right) + \left(x^n - y^n\right) + \left(y^n - y^{n+1} \right)<br />

do for you?
 


Yes, the middle terms cancel like that. But there are generally quite a few of them. I don't get the second line at all. x and y don't have to be integers. I hope you understand this better than you are explaining it.
 


statdad said:
What does

<br /> x^{n+1} - y^{n+1} = \left(x^{n+1} - x^n\right) + \left(x^n - y^n\right) + \left(y^n - y^{n+1} \right)<br />

do for you?

Beats me. How does that help? Just curious.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top