Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ for a natural number $ n \geq 2$? Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ holds for a natural number $ n \geq 2$?
My attempt at solving:
Using $(a+b)^2=a^2+2ab+b^2$; if $(a+b)^2=a^2+b^2$, $2ab=0$ therefore $a$ and/or $b$ must be $0$.
If $a$ and/or $b$ is $0$ then $a^2$ and/or $b^2$ will be $0$.
Therefore $(a+b)^2$ can never equal $a^2+b^2$.

Suppose $a\neq0$. Then
$$(a+b)^n=a^n\Big(1+\tfrac{b}{a}\Big)^n$$
Then, to is enough to consider the question for which values $x$ is $(1+x)^n=1+x^n$. For then $(a+b)^n=a^n+b^n$ with $b=ax$.
This leads to finding all the roots of
$$
p_n(x):=\sum^{n-1}_{k=1}\binom{n}{k}x^k=0
$$
When $n=2$, $p_2(x)=2x=0$ and so $x=0$
when $n=3$, $p_3(x)= 3x+3x^2=0$ and so $x=0$ and $x=-1$,
when $n=4$, $p_4(x)=4x+6x^2 + 4 x^3=2x(2+3x+2x^2)=0$ and so, $x=0$, $x=\frac{-3+i\sqrt{7}}{4}$ and $x=\frac{-3-i\sqrt{7}}{4}$.
and so on.
Notice that pairs $(a, b)$ that satisfy $(a+b)^n=a^n+b^n$ may be complex pairs.
