Motivated by this question.
Let $p\in\Bbb Z[t]$ be a monic polynomial of degree $n$.
Question: When does the equation $p(x)=y^n$ have infinitely many solutions $(x,y)\in\Bbb Z\times\Bbb Z$?.
Some thoughts:
- Using the method in the answer to the linked question we can show that if $n$ is odd this is the case precisely when $p = (t+k)^n$ for some $k\in\Bbb Z$. Indeed, suppose that $p(x)=y^n$ has infinitely many solutions. Write $p = t^n+at^{n-1}+\dots$. Let $k_1,k_2\in\Bbb Z$ be so that $a-nk_1,nk_2-a>0$. By expanding the power and using that $n-1$ is even this implies that $$(x+k_1)^n<p(x)<(x+k_2)^n$$ for almost all $x\in\Bbb Z$. Thus for almost all solutions $(x,y)$ we have $x+k=y$ for some $k_1< k<k_2$. Hence by the pigeonhole principle there is some $k_1<k<k_2$ such that infinitely many solutions are of the form $x+k=y$, i.e. $p-(t+k)^n$ has infinitely many roots, therefore $p=(t+k)^n$.
Maybe there is a similar approach in the case where $n$ is even? - In the special case $n=2$ we can argue as follows that the only possibility is as in the first point, i.e. if $p=(t+k)^2$ for some $k\in\Bbb Z$: If $p=t^2+at+b$, then $p(x)=y^2$ is equivalent to $(2x+a)^2-a^2+4b=y^2$ and this cannot have infinitely many solutions if $-a^2+4b\ne0$ because the distance between distinct square-numbers grows indefinitely (alternatively factor the equation), hence $p=(t+k)^2$ with $k=\frac{a}{2}\in\Bbb Z$.
- We have Siegel's theorem on integral points on smooth curves, so in the case $n>2$ if the equation $p(x)=y^n$ has infinitely many solutions we know at least that $p$ is not squarefree.