I'm trying to show that if $f(x) \in \mathbb{Z}[X]$ is a polynomial of degree $3$. Then there exist infinitely many positive integers $n$ such that $f(n)$ is not a perfect square.

I tried to prove it by splitting into three cases: when $f(x)$ has distinct roots, when $f(x)$ has 1 root of multiplicity $2$, when $f(x)$ has 1 root of multiplicity $3$.

I already proven that the statement is true when $f(x)$ has distinct roots.

enter image description here https://pdfs.semanticscholar.org/5eb7/4363199754164fdf7bbc77925a98c1eb435b.pdf

My attempt: I proved that if $f(x)=x^3+ax+b \in \mathbb{Z}[X]$ is a monic cubic polynomial with a root of multiplicity $2$, then its roots must lie in $\mathbb{Z}$. It follows that there exists infinitely many positive integers $n$ such that $f(n)$ is not a perfect square. I wonder if the same is true for general cubic polynomials, i.e., if $f(x)= \sum_{i=0}^3 a_ix^i$ has a root of multiplicity $2$, will its roots be integers?

  • $\begingroup$ What does multiplicity of the roots have to do with the original question? $\endgroup$ – quasi Nov 29 '19 at 6:53
  • $\begingroup$ I edited the question! $\endgroup$ – yunadesu Nov 29 '19 at 7:05
  • $\begingroup$ Can you show your proof for the distinct roots case? $\endgroup$ – quasi Nov 29 '19 at 7:05
  • $\begingroup$ Can we relate it with Mordell curve $y^2=x^3+n$ as it says that this equation has a finite number of solutions in integers for all nonzero $n$ ( mathworld.wolfram.com/MordellCurve.html )? $\endgroup$ – SARTHAK GUPTA Nov 29 '19 at 7:09

Let $f$ in $\mathbb{Z}[x]$ be such that $\deg(f)$ is odd, equal say, to $2k-1$, for some positive integer $k$.

Let $S$ be the set of all positive integers $n$ such that $f(n)$ is not a perfect square.

Claim:$\;S$ is infinite.

Proof:$\;$Suppose instead that $S$ is finite.

Our goal is to derive a contradiction.

Let $m$ be a positive integer such that $m > n$, for all $n\in S$.

Let $a$ be the leading coefficient of $f$, and let $g(x)=px^2+m$, where $p$ is a prime which doesn't divide $a$.

Then for all integers $n$, we have $g(n)\ge m$, hence $g(n)$ is a positive integer which is not in $S$.

Letting $h(x)=f(g(x))$, it follows that $h(n)$ is a perfect square for all integers $n$.

Hence by the referenced theorem, $h$ is a perfect square in $\mathbb{Z}[x]$.

But then the leading coefficient of $h$ must be a perfect square, contradiction, since the leading coefficient of $h$ is $ap^{2k-1}$, which is divisible by $p^{2k-1}$ but not by $p^{2k}$.

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