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I believe that the answer is yes.

Here's my thinking:

  • $x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$
  • $-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$
  • So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c})$ with $c$ increasing by $1$ at each step and
  • Since $n$ is odd, eventually, we get: $(-1)^c(x^{c}y^{c+1} + x^{c}y^{c+1}) = (x+y)(-1)^c(x^{c}y^{c})$

Am I correct? Am I making any mistakes in my thinking?

If this is correct, is the standard formula:

$$x^n + y^n = (x+y)(x^{n-1} -x^{n-2}y -xy^{n-2} + \dots + (-1)^{\frac{n-1}{2}}x^{\frac{n-1}{2}}y^{\frac{n-1}{2}} + y^{n-1})$$

Thanks,

-Larry

Edit: Fixed the equation based on comment.

Larry Freeman
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    You're about 90% correct, but your 'standard formula' is off at the end - your exponents should be $n$s (not $x$s) and your penultimate one, for instance, should be proportional to $xy^{n-2}$. – Steven Stadnicki Feb 16 '15 at 20:08
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    The answer is yes, your reasoning seems to be fine. It shouldn't be very hard to show it by simple induction. – Galc127 Feb 16 '15 at 20:09

3 Answers3

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Your reasoning is fine.

A more general result is that for any positive integer $n$, and integers $u,v$, $u^n-v^n$ is divisible by $u-v$. This theorem is easier to prove by induction since you are proving for all $n$.

Your theorem then follows when $n$ is odd by setting $u=x,v=-y$ so $u-v=x+y$ and $u^n-v^n=x^n+y^n$.

Thomas Andrews
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$\text{ We have } P(x) = x^n + y^n \to P(-y) = (-y)^n + y^n = -y^n +y^n = 0 \to x+y \text{ is a factor of P(x) }$.

DeepSea
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Hint $\ {\rm mod}\ x\!+\!y\!:\ \color{#c00}{x\equiv -y} \,\Rightarrow\, \color{#c00}x^n+y^n\equiv (\color{#c00}{-y})^n+y^n\equiv 0\,$ by $\,n\,$ odd, and $ $ Congruence Rules.

Community
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Bill Dubuque
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