# Solve $x^n+y^n = (x+y)^n$

Find all positive integers $n$ and real numbers $x$ and $y$ satisfying $x^n+y^n = (x+y)^n$.

We first consider the case that $n$ is even. We have $x^{2k}+y^{2k} = \binom{2k}{0}x^{2k}+\binom{2k}{1}yx^{2k-1}+\cdots+\binom{2k}{2k}y^{2k}$. This can be simplified down to $$\binom{2k}{1}yx^{2k-1}+\binom{2k}{2}y^2x^{2k-2}+\cdots+\binom{2k}{2k-1}xy^{2k-1} = 0 \implies$$ $$\binom{2k}{1}x^{2k}+\binom{2k}{2}yx^{2k-1}+\cdots+\binom{2k}{2k-1}y^{2k} = 0.$$

Similarly we can form equations if $n$ is even. The form of the terms remind me of derivatives so that may be useful.

• Did somebody say.. commutative ring with prime characteristic? Mar 25, 2016 at 20:07
• $n=1$. $x=0$. $y=0$. And $x=-y$ for odd $n$. That's it, right? Mar 25, 2016 at 20:20
• For $x \neq 0,$ let $y = r |x|.$ Either you have $x > 0,$ $1 + r^n = (1+r)^n,$ different cases $n$ odd/even, or $x < 0,$ $-1 + r^n = (-1+r)^n.$ Draw some graphs. Mar 25, 2016 at 20:29
• If $n$ is even, then either $x=0$ or $y=0$, and if $n$ is odd, then $x=-y$ or $x=0$ or $y=0$.
– user217174
Mar 25, 2016 at 20:47
• @2000 So there are no solutions? Mar 25, 2016 at 20:51

$y=0$ is a solution for every $n>0$, so we can assume $y\ne0$. For $n=1$ the equation is satisfied for every pair of numbers $x$ and $y$; thus we can assume $n>1$.

If we set $x=ty$, the equation becomes $$y^n(t^n+1)=y^n(t+1)^n$$ so we can as well solve $t^n+1=(t+1)^n$.

Consider the function $f(t)=(t+1)^n-t^n-1$, which we want to find the zeros of. We have $$f(t)=\sum_{k=1}^{n-1}\binom{n}{k}t^k$$ so $$\lim_{t\to\infty}f(t)=\infty$$ whereas $$\lim_{t\to-\infty}f(t)=\begin{cases} -\infty & \text{if n is even}\\ \infty & \text{if n is odd} \end{cases}$$ For the derivative $$f'(t)=n((t+1)^{n-1}-t^{n-1})$$ we see it vanishes for $$\left(\frac{t+1}{t}\right)^{\!n-1}=1$$

If $n$ is even, $n-1$ is odd and so the derivative does not vanish. The function has then a single zero.

If $n$ is odd, the derivative vanishes for $t+1=-t$, that is, $t=-1/2$, which is the point of absolute minimum. Since $$f(-1/2)=(1/2)^n-(-1/2)^n-1=\frac{1}{2^{n-1}}-1<0$$ the function has two zeros.

Note that $f(0)=0$ for every $n>1$; if $n$ is odd, $f(-1)=0$.

Summarizing the above facts, we have that the equation $(x+y)^n=x^n+y^n$, for $n>1$, has the trivial solutions “$x=0$ or $y=0$”. If $n$ is odd the equation has also the solution $x=-y$.

No other solutions exist for $n>1$.