Question: Under what circumstances/restrictions on $x$ and $y$ does $(x + y)^n = x^n + y^n$ given the value of $n$? That is, what can we tell about $x$ and $y$ from the value of $n$ and the equation $(x + y)^n = x^n + y^n$? I'd be satisfied with $ x, y \in \mathbb{Z} $ and $ n \in \mathbb{N} $.

Background: Spivak's Calculus' Question 1-16 asks us to find the restrictions individually from $n = 1$ till $n = 5$; and invited readers to guess at the general pattern for any $n$, which I still couldn't see.

Research: This is a subset of Sums of powers being powers of the sum where $ \alpha = \beta $ and $ k = 2 $; but I want a more specific and simpler explanations.

Edit: The original, ambiguous framing of this question was interpreted by commenters to mean asking about the limitations on $n$ to satisfy $ (x + y)^n = x^n + y^n $ for any $x$ and $y$. I have since edited it to (hopefully) better reflect my original intent; and apologize to the answerers.

  • $\begingroup$ $n=1$? ${}{}{}{}$ $\endgroup$ – evil999man May 10 '14 at 10:31
  • $\begingroup$ Do you mean "would the equation ... be true for every $n \in N$"? $\endgroup$ – barak manos May 10 '14 at 10:51
  • 1
    $\begingroup$ This might be interesting. $\endgroup$ – user1337 May 10 '14 at 11:03
  • $\begingroup$ @YatharthROCK Obviously, for the conditions you stated n=1 is the only solution. $\endgroup$ – evil999man May 10 '14 at 12:57
  • $\begingroup$ @Awesome: Aah... thanks, I finally realize the discrepancy between my intent of asking the question and how it was originally framed (see my edit). $\endgroup$ – Yatharth Agarwal May 10 '14 at 14:15

$$(x+y)^n=x^n+y^n$$ By the Binomial Theorem we can expand the left side to $$\sum_{k=0}^{n}{\binom{n}{k}}x^{n-k}y^k=x^n+y^n$$ Isolate the $x^n$ and $y^n$ by changing the bounds on the summation. $$x^n+\sum_{k=1}^{n-1}{\binom{n}{k}}x^{n-k}y^k+y^n=x^n+y^n$$ Thus the two sides are equal when $$\sum_{k=1}^{n-1}{\binom{n}{k}}x^{n-k}y^k=0$$ Notice that, if $x,y\in \mathbb{N}$, $\sum_{k=1}^{n-1}{\binom{n}{k}}x^{n-k}y^k$ is strictly positive and only $n=1$ is a solution. Relaxing to $x,y\in\mathbb{Z}$ give more possibilities of solutions. For example, if $x=1, y=-1$, then the equation is satisfied for $n=2m-1$, where $m\in \mathbb{N}$


It's easy to show from the binomial theorem, that for every $n\neq1$, the equation has solutions only if $x=0\lor y=0$, because otherwise we come up with additional not-null terms (and binomial terms are always positive)

(That assuming $n\in\mathbb{N}$, however that question would be much more interesting if $x,y,n \in \mathbb{R}$, but I wouldn't have the necessary knowledge to answer you. Indeed I suspect the solution is the same...)


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