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

$y=0$ 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 $n>1$.