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.
 A: $y=0$ 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 $n>1$.
