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Recently I have been interested in the problem of when a polynomial $q \in \mathbb{Z}[x]$ evaluates to $k$th powers (for some fixed $k \geq 2$) for all inputs $x \in \mathbb{N}$. The claim is all such polynomials are themselves $k$th powers in $\mathbb{Z}[x]$. An outline of the proof of this for $k=2$ is as follows:

-We can reduce to the case where $q$ has distinct irreducible factors.

-Given any integer polynomial we have infinitely many primes $p$ such that $p | q(n)$ for some $n \in \mathbb{N}$.

-We have $p$ divides $q(n)$ implies $p^2$ divides $q(n)$ since $q$ is square valued.

-$q(n+tp)$ is divisible by $p$ and hence divisible by $p^2$.

-$q(n+tp) \equiv q(n) + t p q'(n) \pmod{p^2}$

-$\textbf{By picking $t$ coprime to $p$ we see $p$ divides $q'(n)$.}$

-Then if we look at the ideal $I = (q,q') \subset \mathbb{Z}[x]$ we see $I \cap \mathbb{Z} = 0$.

-Then $q$ and $q'$ share a root, so $q$ has repeated roots, a contradiction.

All parts of the outline generalize to the case $k > 2$ except for the bold step. I was hoping for some advice on how to extend this step.

$\textbf{Edit to clarify:}$ In the $k > 2$ case we get $0 \equiv q(n + t p) \equiv \sum_{j=0}^{k-1} (t p)^j \frac{q^{(j)}(n)}{j!} \pmod{p^{k}}$. Then we need to show this implies $q^{(j)}(n)$ is divisible by $p$ for all $j$.

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The bold step follows directly from the preceding steps: since $p^2$ divides both $q(n+tp)$ and $q(n)$, then $p^2 | tpq'(n)$ which implies $p|t q'(n)$ which implies $p | q'(n)$ (if $t$ is chosen prime to $p$).

I am actually wondering about the step after the bolded step, which I don't understand. Certainly for an irreducible polynomial $q(x) \in \mathbb{Z}[x]$, we know that $q(x)$ and $q'(x)$ are relatively prime in $\mathbb{Q}[x]$. So we can choose $f(x), g(x) \in \mathbb{Q}[x]$ such that $f(x) q(x) + g(x) q'(x) = 1$. Clearing denominators from $f$ and $g$, we conclude that the $\mathbb{Z}[x]$-ideal $(q,q')$ contains a non-zero integer. This is the opposite of what is claimed. So somehow the contradiction must come from the preceding steps in the proof, but I don't see it.

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    I understand how it follows from the previous step for the $k=2$ case. I was wondering how it works for $k > 2$ (here the previous step is more than a linear polynomial). For your concern: $q(x)$ is not irreducible, it is simply squarefree in $\mathbb{Z}[x]$. As you noted $(q(x),r(x)) \cap \mathbb{Z} = \{0\}$ if and only if $q$ and $r$ share a factor. – 2017-01-04
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    I don't understand what you mean by "the previous step is more than a linear polynomial". All the previous steps are valid for $k>2$, and the bold step follows from the previous steps (which don't mention $k$ at all). – 2017-01-04
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    I have edited to hopefully clarify the confusion for $k > 2$. Since I am trying to show for $k > 2$ that the polynomial is a $kth$ power, I need to show the prime divides $q(n), \ldots, q^{(k-1)}(n)$. – 2017-01-04