# Is this problem of Schinzel and Tijdeman misquoted? It appears easy with Pell equations

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 $z$?

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.

Is the problem misquoted?

• If you google acta arithmetica 1976, guess what you will find. – Franz Lemmermeyer Dec 6 '15 at 10:42
• @FranzLemmermeyer Many thanks! I already found the same paper, but as far I can tell, the title is different from the cited, which confused me. – joro Dec 6 '15 at 11:11

The problem is indeed misquoted. Here is the correct statement:

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 $yz\neq 0$.

The paper can be found here.

• Thanks, I saw this in another paper. Is it possible this to be another problem? One of the polynomials is over integers, the is over rationals. – joro Dec 6 '15 at 9:28
• The paper by Schinzel and Tijdeman given as a reference to the problem you ask about only mentions one conjecture, which is this one. – Wojowu Dec 6 '15 at 9:31
• I edited, adding the paper. Are you quoting the same paper? – joro Dec 6 '15 at 9:41
• Is the paper with the same title? (I found another title in the journal). Maybe mistake of the citer. – joro Dec 6 '15 at 12:56
• Indeed, the title is "On the equation $y^m=P(x)$", by A. Schinzel and R. Tijdeman, from Acta Arithmetica XXXI (1976). – Wojowu Dec 6 '15 at 13:14