<s>[INST] Is $g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)} $ always an integer? Let \begin{equation} f(x, y) = \frac{x^y-1}{x+1} \end{equation}
And  \begin{equation}
g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}
\end{equation}
where $x, y $ are positive integers with $y$ even, $f(x, y) \not = 1$.
It appears $g(x, y) $ is always an integer.
To prove this, I tried fixing $x$ and doing induction on $y$ but got stuck in the induction step. (An elementary proof would be excellent)
 [/INST] Here is an elementary proof:
Let us change the notation so that
$$ f(x,y):= \frac{x^{2y}-1}{x+1}$$ and
$$ g(x,y):= \frac{f(x,y)^{2y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}.$$
We first show that $f(x,y)-1$ and $xf(x,y)+1$ are coprime.  Suppose that there exist $p$ a prime which divides both $f(x,y)-1$ and $xf(x,y)+1$ then $p$ divides $(x+1)f(x,y)$ then $p$ divides $x+1$.
Then if $p$ divides $f(x,y)-1$ then $p^2$ divide $x^{2y+1}-x-2$, and if $p$ divides $xf(x,y)+1= \frac{x^{2y+1}+1}{x+1}$ then $p^2$ divides $x^{2y+1}+1$. Then $p^2$ divides $x+3$.
Then $p$ must be equal to $2$ and $x$ must be odd, but then $xf(x,y)+1= 1-x+x^2- \cdot\cdot +x^{2y}$ must be odd, which is contradiction.
Now if $q$ is prime and $q^h$ , $h \gt 0$, is the largest power of $q$ which divides $f(x,y)-1$ then since $f(x,y)-1$ divides $f(x,y)^{2y+1}-1$ ,then $q^h$ also divides $f(x,y)^{2y+1}-1$.
And if $q^h$ , $h\gt 0$, is the largest power of $q$ which divides $xf(x,y)+1$ then $-xf(x,y) \equiv 1 \bmod q^h$, then  neither $x$ nor $f(x,y)$ is divisible by $q$ and $f(x,y)^{2y+1}-1 \equiv -\frac{1}{x^{2y+1}}-1 \equiv   -\frac{x^{2y+1}+1}{x^{2y+1}} \bmod q^h$. That is
$$ x^{2y+1} (f(x,y)^{2y+1}-1) \equiv - (x+1) (xf(x,y)+1) \equiv 0 \bmod q^h$$ then $q^h$ divides $f(x,y)^{2y+1}-1$.
</s>