if $(k+m)^n + (l+m)^n = (k+l+m)^n$ then $kl$ divides $m^n$

Question

How to prove that for any $k, l, m, n \in \Bbb N$ and $n \ge 3 $ with $(k+m)^n + (l+m)^n = (k+l+m)^n$ then $kl$ divides $m^n$?

As a special case, in the same way as Brian's answer below it can be shown that if $a^n + b^n = (a+k)^n$ then $k$ divides $b^n$: $$a^n + b^n = (a+k)^n = a^n + k^n + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i}\\ = a^n + k^n + k\sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1} \\= a^n + k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$ so $$ b^n = k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$ thus $k$ divides $b^n$ and furthermore $ rad(k) $ divides $b$.

Answer 1

By Fermat's last theorem (yes, not Fermat's little theorem) one of $k+m$ and $l+m$ is zero, which means that $m=0$ and one of $k$ and $l$ is zero. It follows that $kl=0$ and $m^n=0$; $0\mid0$ and the statement is true.

Answer 2

Use the binomial theorem to expand both sides of the original equation:

$$\sum_{i=0}^n\binom{n}i\left(k^i+\ell^i\right)m^{n-i}=\sum_{i=0}^n\binom{n}i(k+\ell)^im^{n-i}\,.$$

Pull out the $m^n$ terms:

$$2m^n+\sum_{i=1}^n\binom{n}i\left(k^i+\ell^i\right)m^{n-i}=m^n+\sum_{i=1}^n\binom{n}i(k+\ell)^im^{n-i}\,,$$

so

$$\begin{align*} m^n&=\sum_{i=1}^n\binom{n}im^{n-i}\left((k+\ell)^i-\left(k^i+\ell^i\right)\right)\\ &=\sum_{i=1}^n\binom{n}im^{n-i}\sum_{j=1}^{i-1}\binom{i}jk^j\ell^{i-j}\\ &=k\ell\sum_{i=1}^n\binom{n}im^{n-1}\sum_{j-1}^{i-1}\binom{i}jk^{j-1}\ell^{i-j-1}\,. \end{align*}$$