<s>[INST] 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.
 [/INST] 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 $(1 + m^n, m \cdot (1 + m^n), 1 + m^n)$.
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