# cubic polynomial with a root of multiplicity 2

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.

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?

• What does multiplicity of the roots have to do with the original question? – quasi Nov 29 '19 at 6:53
• I edited the question! – yunadesu Nov 29 '19 at 7:05
• Can you show your proof for the distinct roots case? – quasi Nov 29 '19 at 7:05
• 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 )? – 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 $$p^{2k}$$.