Related to FLT and this question.
For natural $n > 4 $ define the curve $C_n : z^{n-2}y(y+z)=x^n$.
$C_n$ has the trivial points with $x=0$ for all $n$.
The answer in the linked question shows there are no other solutions using FLT.
I am interested in elementary approach without using FLT or modularity.
Q1 Do we get constraints or lack of non-trivial coprime integral points $(X,Y,Z)$ on $C_n$ via elementary approach not using modularity?
Partial results, could be wrong. Assume $(X,Y,Z)$ is coprime point.
If prime $p$ divides $Z$ or $Y$ then it divides $X$ too.
Assume $p \mid Y$ and $p^2 \nmid Y$. Since $p \mid X$ then $p$ is coprime to $Z$. Set $Y=p Y_1, X=p X_1$. Then $Z^{n-2} p Y_1 (p Y_1 + Z)=(p X_1)^n=p^n X_1^n$. The LHS is divisible by $p$ and is not divisible by $p^2$ and the RHS is divisible by $p^n$ and we can't equate the exponents of $p$.
Likewise, assume $p \mid Z$ and $p^2 \nmid Z$. Since $p \mid X$ then $p$ is coprime to $Y$. Set $Z=p Z_1, X=p X_1$. Then $p^{n-2} Z_1^{n-2} Y (Y + p Z_1)=(p X_1)^n=p^n X_1^n$. The LHS is divisible by $p^{n-2}$ and is not divisible by $p^{n-1}$ and the RHS is divisible by $p^n$ and we can't equate the exponents of $p$.
Q2 Is this correct proof that the non-trivial points are with $Y,Z$ powerful integers?
