In [Diophantine equations over the twentieth century: a (very) brief overview
, p. 5](http://www-irma.u-strasbg.fr/~bugeaud/travaux/prockyoto.ps)



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> **Problem** Let $f(x) \in \mathbf{Z}[x]$ be an irreducible polynomial of degree at least 2. Do the Diophantine equation $$f(x)=y^2z^3$$ have only finitely many solutions in non-zero integers $x,y$ and $z$?
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The most trivial case is $z=1,f(x)=ax^2+1$ where $a$ is not square.


This leads to the Pell equation $y^2-ax^2=1$, which has infinitely many solutions.


Another approach is let $f(x)=x^2+1$. For fixed $z$, this leads to Pell equation
$x^2-z^3 y^2= -1$. For infinitely many $z$, it has infinitely many solutions $x,y$.


Couldn't find the reference "[13] A. Schinzel and R. Tijdeman, On the equation $y^ m = f(x)$, Acta Arith. 31 (1976),
199-204." online.



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> Is the problem misquoted?
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Is this problem of Schinzel and Tijdeman misquoted? It appears easy with Pell equations
The problem is indeed misquoted. Here is the correct statement:



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> If a polynomial $P(x)$ with rational coefficients has at least three simple zeros then the equation $y^2z^3=P(x)$ has only finitely many solutions in integers $x,y,z$ with $yz\neq 0$.
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The paper can be found [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa31/aa3129.pdf).


