# $f (x) \in\mathbb Z[x]$ be such that $f$ is a perfect $k$-th power on taking positive integer values . Is $f$ a perfect $k$-th power?

Let $f(x)\in\mathbb Z[x]$ and integer $k>1$ such that $f(n)$ is a perfect $k$-th power for every positive integer $n$. Is it true that there is a $g(x)\in\mathbb Z[x]$ such that $f(x)=\left(g(x)\right)^k$ ?

If not , then does the stronger assumption , $f(n)$ is a perfect $k$-th power for every integer $n$ implies there is a $g(x)\in\mathbb Z[x]$ such that $f(x)=\left(g(x)\right)^k$ ?

Perhaps there exist a simpler answer.(I think that what follows is in Polya-Szego, but I am not sure).

We work first in $\mathbb{Q}[x]$.

1) Suppose first that we have $f(x)=c^k x^{mk}+\cdots$. Define $\displaystyle h_1(x)=\frac{f(x)}{c^k x^{mk}}=1+\frac{a_1}{x}+\cdots+\frac{a_{mk}}{x^{mk}}$. We have that all $a_k$ are in $\mathbb{Q}$.

Put now $\displaystyle h(x)=\sqrt[k]{h_1(x)}=1+\frac{b_1}{x}+\cdots+\frac{b_{j}}{x^{j}}+\cdots$. $h(x)$ is a power series in the powers of $1/x$, again with coefficients in $\mathbb{Q}$, converging for large $|x|$.

Put $$g(x)=cx^{m}h(x)=\sum_{j=0}^{m}cb_jx^{m-j}+\sum_{j\geq m+1}\frac{cb_j}{x^{j-m}}=P(x)+Q(x)$$

Note that $Q(x)\to 0$ if $|x|\to +\infty$.

Now let $x=n$ a large integer. As $f(n)=g(n)^k$ and $g(n)\in\mathbb{R}$, we get that $g(n)\in \mathbb{Z}$. Let $d$ in $\mathbb{N}$ such that $dP(x)\in \mathbb{Z}[x]$. We have hence $dg(n)=dP(n)+dQ(n)$, hence $dQ(n)\in \mathbb{Z}$. If $n$ is sufficiently large, we have $d|Q(n)|<1$, hence $Q(n)=0$ and $g(n)=P(n)$. We have $f(n)=(P(n))^k$ for large $n$, and hence $f(x)=(P(x))^k$.

2) Now we suppose only that $f(x)=c x^m+\cdots$. Choose $q_1,\cdots, q_k$ large integers such that for all $i$ $f(x+q_i)$ is prime (in $\mathbb{Q}[x]$) to $\prod _{j\not =i}f(x+q_j)$. Then $f_1(x)=\prod f(x+q_i)$ verify the hypothesis, and is of the form $f_1(x)=c^kx^{mk}+\cdots$. By the above, $f_1$ is the $k$-th power of a polynomial. Then each $f(x+q_j)$ is also, up to a constant, of the same form. Hence there exists a constant $d$, and a polynomial $H$, such that $f(x)=d(H(x))^k$. Taking for $x$ an integer such that $H(x)\not =0$, the hypothesis show that $d$ is the $k$-th power of a rational, and finally, we can write $f(x)=(P(x))^k$, with $P\in \mathbb{Q}[x]$.

3) It remain to show that $P\in \mathbb{Z}[x]$. There exists $d\in \mathbb{N}$ such that $dP(x)=\sum c_j x^j=P_1(x)$, with the $c_j \in \mathbb{Z}$ relatively prime. The equality $d^kf(x)=(P_1(x))^k$ show that $d=1$, and we are done.

• @Thomas Andrews Thanks a lot, I have edited. – Kelenner Dec 4 '15 at 19:28

Let $f\in\Bbb{Z}[x]$ with roots $\alpha_1,\ldots,\alpha_d\in\Bbb{C}$ and multiplicities $m_1,\ldots,m_d$, so that $$f=c\cdot\prod_{i=1}^d(x-\alpha_i)^{m_i},$$ for some $c\in\Bbb{Z}$. For each root there are infinitely many primes that split completely in $\Bbb{Q}(\alpha_i)$. Choose such a prime $p_i$ for every $\alpha_i$, in such a way that every $p_i$ is coprime to $c$ and to $\alpha_i-\alpha_j$ for all distinct $i$ and $j$. Then by the Chinese remainder theorem there exists an integer $n$ such that $$n\equiv\alpha_i\mod p_i\qquad\text{ and }\qquad n\not\equiv\alpha_i\mod p_i^2.$$ In this way, the $p_i$-adic valuation of $f(n)$ is precisely the multiplicity $m_i$ of $\alpha_i$ for every $i$. If $f(n)$ is a perfect $k$-th power then each of the $m_i$ must be a multiple of $k$, say $m_i=m_i'k$, so that $$f=c\prod_{i=1}^d(x-\alpha_i)^{m_i'k}=c\left(\prod_{i=1}^d(x-\alpha_i)^{m_i'}\right)^k,$$ and because $f(n)$ is a perfect $k$-th power also $c$ is a perfect $k$-th power, say $c=c'^k$. This shows that $f$ is a $k$-th power of some $g\in\Bbb{C}[x]$, and of course $g\in\Bbb{Z}[x]$ because $g^k\in\Bbb{Z}[x]$.

Therefore, if $f(n)$ is a perfect $k$-th power for some such $n$, then $f=g^k$ for some $g\in\Bbb{Z}[x]$.

• "Coprime" needs to be carefully worded to apply CRT. And you haven't stated in which ring, unless by $\mathbb Q(\alpha_i)$ you mean $\mathbb Q(\{\alpha_i\})$. Finally, of course, the $\alpha_i$ are not necessarily algebraic integers, although that can be taken care of, I believe. – Thomas Andrews Dec 4 '15 at 19:02
• @ThomasAndrews You are right, I'll edit to clarify. – Inactive - Objecting Extremism Dec 4 '15 at 19:04

HINT: Let's show that the function $x \mapsto h(x) = \sqrt[k]{f(x)}$ is a polynomial with rational coefficients. First, it takes only natural values for $n$ large enough. Second, taking the divided difference $\Delta^N h(x) \to 0$ for $x \to \infty$, provided that $N > \frac{\deg f}{k}$. Since the values of the sequence $\Delta^N h(n)$ are integers for large enough $n$, it follows that $\Delta^N h(n)=0$ for large enough $n$. It follows that there exists a polynomial with rational coefficients that gives $h(n)$ for large enough $n$ ( so for all $x$).

• Shouldn't it be $h(x)=(f(x))^{1/k}$ ? Why $\Delta^N h(x) \to 0$ as $x \to \infty$ ? Why are the values of $\Delta^N h(n)$ integers for large $n$ ? Could you please elaborate – user228168 Dec 5 '15 at 7:55