infinitely many solutions to $\displaystyle x^n + y^n = z^{n+1}$ Let $n$ be a positive integer.  Show that the equation
$\displaystyle x^n + y^n = z^{n+1}$
has infinitely many integer solutions.
 A: Hint: It is pretty easy to find infinitely many with $x|y$.
A: Theorem : $x^n+y^n=z^{n+1}$
Let $p$ is an integer and $q$ is another integer.
\begin{equation}
  \left\{
      \begin{aligned}
        &p^n+q^n = k\\
        &OR\\
        &(p^n+q^n)(p^n+q^n)^n= k^{n+1}\\
        &OR\\
        &(p(p^n+q^n))^n+(q(p^n+q^n))^n= k^{n+1}\\
      \end{aligned}
    \right.
\end{equation}
Let $p(p^n+q^n)=x$ and $q(p^n+q^n)=y$ and $k=z$
$x^n+y^n=z^{n+1}$ (Proved)
A: Let $x$ and $y$ be such that $x \mid y$.  This means that $y = x \cdot m$ where $m$ is a positive integer.
Thus
$x^n + y^n = x^n + (x \cdot m)^n = x^n + x^n \cdot m^n = x^n \cdot [1 + m^n] = z \cdot z^n = z^{n+1}$
So we know that when $x \mid y$, we have $x^n \cdot [1 + m^n] = z^n \cdot z$
Now let $x^n = z^n$.  Thus $1 + m^n = z$
Then $z^n = [1 + m^n]^n = x^n$.
Substituting $[1 + m^n]^n$ for $x^n$ we get
$$x^n + y^n = [1 + m^n]^n + (m[1 + m^n])^n = [1 + m^n]^n + m^n \cdot [1 + m^n]^n
 \\ = [1 + m^n] \cdot [1 + m^n]^n = [1 + m^n]^{n+1} = x^{n+1} = z^{n+1}$$
Since there are infinitely many choices for $m$ since $n$ is a positive integer, we have our answer;  a triple that gives infinitely many solutions to the equation
$x^n + y^n = z^{n+1}$
is $(1 + m^n, m \cdot (1 + m^n), 1 + m^n)$.
