For positive integer $n$, let $f_n(x,y,z)＝x^n＋y^n-α^n-β^n$
$α＝\frac{x＋y＋\sqrt{(x＋y)^2-4z^2}}2$, $β＝\frac{x＋y-\sqrt{(x＋y)^2-4z^2}}2$
Then, it is known that $f_n(x,y,z)∈\Bbb{Z}[x,y,z]$. My question is, Why $f_n(x,y,z)$ can be divided by $(xy-z^2)$ ? Heavy calculation and induction may get answer, but other good ideas? Thank you for your help.
Note that $p^{n+1}+q^{n+1}=(p+q)(p^n+q^n)-pq(p^{n-1}+q^{n-1})\tag{$\star$}$
Using the above, along with noting $\alpha, \beta$ are obviously roots of $t^2-(x+y)t+z^2$, we have by Vieta (or by direct calculation), $$\alpha^{n+1}+\beta^{n+1}=(x+y)(\alpha^n+\beta^n)-z^2(\alpha^{n-1}+\beta^{n-1})$$ $$\implies x^{n+1}+y^{n+1}-f_{n+1} = (x+y)\left(x^n+y^n-f_n \right)-z^2\left(x^{n-1}+y^{n-1}-f_{n-1} \right)$$ Clubbing terms which do not involve $f$ and using $(\star)$ on them, $$\implies f_{n+1} = (z^2-xy)\left(x^{n-1}+y^{n-1}\right)+(x+y)f_n -z^2f_{n-1}$$ $$\implies f_{n+1} \equiv (x+y)f_n-z^2f_{n-1} \pmod{xy-z^2}$$ As $f_0=f_1=0$, this recursion inductively shows $\forall n \in \mathbb Z^+, \: f_n \equiv 0 \pmod{xy-z^2}$.