Let $f(x)$ be a non-constant polynomial with integer coefficients. It is a well-known result that if $f(n)$ is a square for all integers $n$, then $f$ must in fact be the square of a polynomial (see, for example: http://www.mast.queensu.ca/~murty/poly2.pdf).

My question is the following. Suppose that $\deg f = d$, and suppose that for every integer $n$, we have that $f(n)$ is a perfect $k$-th power for some $k_n > 1$ dividing $d$. I want to emphasize that $k_n$ is allowed to depend on $n$. For instance, it could be the case that $f$ is of degree $6$ and $f(2) = 2^3$ while $f(3) = 3^2$, and $f(n)$ is either a square or a cube (or both) for every $n$.

Can we conclude that $f$ is a perfect $m$-th power, for some $m > 1$ dividing $d$?

  • $\begingroup$ Yes, of course. See Davenport, Lewis, Schinzel, Polynomials of certain special type, Acta Arithmetica IX, 1964 - or Schinzel's collected works. $\endgroup$ Nov 30 '15 at 3:44
  • $\begingroup$ I saw the paper which states your claim in the first paragraph, but that is not what I wanted to know about. I apologize for the vagueness of the original question. $\endgroup$ Nov 30 '15 at 3:48
  • 3
    $\begingroup$ OK, then there is a $k > 1$ (dividing $d$ as you require) which works for infinitely many $n$ - and you may apply Siegel's theorem (finiteness of integral points of an irrational affine algebraic curve - in this case, the components of $y^k = f(x)$), with the conclusion that $f$ is a $k$-th power. If you want a more "elementary" proof, my best guess is the DLS argument applies just as well in your situation. $\endgroup$ Nov 30 '15 at 3:56
  • $\begingroup$ @VesselinDimitrov Thanks, that argument is perfectly fine for me! $\endgroup$ Nov 30 '15 at 4:04
  • 4
    $\begingroup$ Why the downvote? The question seems reasonable to me. $\endgroup$ Nov 30 '15 at 8:06

Here is a more elementary argument than that of nfcd23. It only assumes that $f(n)$ is a $k$th power for some $k$ depending on $n$, which need not be assumed to divide $d$. Over $\mathbf{C}$, write

$$f(x) = C \prod (x - \alpha_i)^{r_i},$$

where the $\alpha_i$ are distinct. By Cebotarev, (or more simply, by Frobenius), we may find infinitely many primes $p$ such that $p$ splits completely in $K = \mathbf{Q}(\alpha_i)$, and in addition, $p$ is prime to both $C$ and the difference $\alpha_i - \alpha_j$ of any pair of distinct roots. (The last two conditions hold automatically for all but finitely many primes.)

For each root $\alpha_i$, choose a prime $p_i$ of this form. Then choose an integer $n$ (using the Chinese Remainder Theorem) such that:

$$\text{$n$ is congruent to $\alpha_i$ modulo $p_i$ but not modulo $p^2_i$.}$$

This ensures that the $p_i$-adic valuation of $f(n)$ is $r_i$ for each $i$, because that is the valuation of $(n - \alpha_i)^{r_i}$, and by assumption, $p_i$ does not divide any of the other terms. If $f(n)$ is a perfect $k$th power, this implies that the greatest common divisor $r$ of all the $r_i$ must be divisible by $k$, and hence equal to $mk$ where $k > 1$. But that implies that $f(x) = A g(x)^{mk}$ for some $g(x)$. Once again letting $x = n$ and noting that $f(n)$ is a $k$th power, we deduce that $A = B^k$, and so

$$f(x) = B^k g(x)^{mk} = (B g(x)^m)^k.$$

  • $\begingroup$ It's not clear to me if this argument is more elementary because it uses $L$ functions, but I'm glad you solved the more general version. $\endgroup$
    – Will Sawin
    Nov 30 '15 at 8:58
  • 6
    $\begingroup$ @WillSawin: This proof is much more elementary after noting that it is not necessary to use Cebotarev (or Frobenius) here: Let $F(x)\in\mathbb Z[x]$ be the minimal polynomial of an integral primitive element of the splitting field of $f(x)$. Then for any prime $p$ dividing $F(m)$ for some integer $m$ for which $F(x)$ is separable modulo $p$ we get that $F(x)$ factors into linears modulo $p$, and so does $f(x)$. Running through $m\in\mathbb Z$, we get infinitely primes of the required form. $\endgroup$ Nov 30 '15 at 10:35

Here is an argument avoiding Siegel's theorem (and also seemingly different from the DLS argument to which Dimitrov refers, as the main tool I will use is Weil's RH for absolutely irreducible curves whereas DLS use arguments based on Hilbert irreducibility).

Let $S$ be a non-empty finite set of primes (e.g., the primes factors of $d$) and consider $f \in \mathbf{Z}[X]$ that is not an $e$th power in $\mathbf{Z}[X]$ (equivalently, in $\mathbf{\mathbf{Q}}[X]$) for each $e \in S$. An elementary argument with the monic multiple of $f$ shows that for each of the primes $e \in S$ the polynomial $f$ either (i) is not an $e$th power in $\overline{\mathbf{Q}}[X]$ or (ii) is of the form $c h^e$ for some monic $h$ with $c$ the leading coefficient of $f$. If there is any integer $n$ away from zeros of $f$ such that $f(n)$ is an $e$th power for some $e$ as in case (ii) then $c$ is an $e$th power and hence so is $f$. Thus, we can assume for our purposes that case (ii) never occurs.

Since each $e\in S$ is prime, it follows that the polynomial $Y^e - f(X)$ is irreducible in $\overline{\mathbf{Q}}[X,Y]$ (as it is the same to be irreducible in $\overline{\mathbf{Q}}(X)[Y]$, for which $f$ not being an $e$th power in $\overline{\mathbf{Q}}(X)$ is equivalent to the irreducibility property since $e$ is prime). We conclude that $Y^e - f(X)$ is absolutely irreducible over $\mathbf{Q}$ for every $e \in S$. Hence, for all large primes $p$, $Y^e - f(X)$ is absolutely irreducible over $\mathbf{F}_p$ too. (Recall that "absolute irreducibility" is inherited under reduction modulo all but finitely many primes, whereas ordinary irreducibility is not.) In what follows, only consider such $p$ (moreover big enough so that $f \bmod p$ has the same degree as $f$).

These curves $Y^e - f(X) = 0$ for varying $e \in S$ (if $\#S > 1$) have finite overlap in characteristic 0, so they are pairwise disjoint up to uniformly bounded error in characteristic $p$ for large $p$. Hence, the solution set $V_p$ to $h \equiv 0 \bmod p$ is the "disjoint" (up to bounded amount) union of the solutions sets to the individual curves $C_{e,p} := \{y^e - f(x) = 0\}$. There are at most $d := \deg(f)$ values $x \in \mathbf{F}_p$ where $f$ vanishes mod $p$, over which there is only one point $(x,0)$ in $V_p$. Ignoring those at most $d$ points, as well as the uniformly bounded overlaps sets for distinct $e$'s just mentioned, every other fiber of $V_p$ over the $x$-line $\mathbf{F}_p$ lies in exactly one of the curves $C_{e,p}$.

Consider $p \equiv 1 \bmod e$ for all $e \in S$. The fibers for $C_{e,p}$ have size $e$ (away from zeros of $f$ in $\mathbf{F}_p$). As $p$ grows, $\#C_{e,p}(\mathbf{F}_p) \sim p$ for each $e \in S$, by RH, so the image of $C_{e,p}$ in $\mathbf{F}_p$ consists of $\sim p/e$ points as $p$ grows.

Varying through all $e \in S$, if $V_p$ actually hits the entire $x$-line for all such large $p$ (say even up to a bounded amount as such $p$ grows) we would get $\sum_{e \in S} 1/e = 1$ (equality on the nose, not just approximation). But the $e$'s are pairwise distinct primes, so no such equality is possible (look at it $e_0$-adically for one $e_0 \in S$).

Thus, for (many) large $p$ the projection $x: V_p \rightarrow \mathbf{F}_p$ is not surjective, so if $n \in \mathbf{Z}$ represents a mod-$p$ residue class not in the image then for every $e \in S$ the congruence $y^e \equiv f(n) \bmod p$ has no solution, so certainly for every $e \in S$ the integer $f(n)$ is not an $e$th power in $\mathbf{Z}$ as well.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.