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We know following theorem by Schur: Suppose that $f(x) \in \mathbb{Z}[x]$ is a polynomial such that exists an integer $m$ such that $f(n) = m^2$ for every integer $n$. Then $f(x) = g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.

Is this true if I using $3,4,5 \cdots$ instead of $2$?

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Should we assume that your $m$ is not fixed? Meaning, is there a single $m \in \mathbb{Z}$ such that $f(n) = m^2$ for all integers $n$? If so, $f(x) := m^2$. –  Benjamin Dickman Oct 28 '12 at 17:45
    
Yes it is true. See Cor. 3.3 in isibang.ac.in/~sury/polys.pdf –  J.C. Ottem Oct 28 '12 at 18:19
    
(assuming the quantifiers are 'for every $n$ there exists and $m$ such that $f(n)=m^2$).. –  J.C. Ottem Oct 28 '12 at 18:23
    
The same finite-difference trick that solves the square-root version of this problem (MO#110904) works here too. –  Noam D. Elkies Oct 28 '12 at 18:33

2 Answers 2

up vote 2 down vote accepted

I think the proposer tried to say that if for every integer $n$ there is an integer $m$ such $f(n)=m^2$, then $f(x)=g(x)^2\dots$.

The generalization from exponent $2$ to higher exponents $e$ is an exercise in the second volume of the classical Polya/Szegö problem book. I don't have it at hand now, so I cannot give a more precise reference.

If one allows more sophisticated tools, then it follows immediately from Hilbert's irreducibility theorem: Let $t$ be another variable, and factorize $f(x)-t^e$ over $\mathbb Q(t)$ into irreducible factors $P_i(t,x)$. By Hilbert, there are infinitely many integers $n$ such that $P_i(n,x)$ is irreducible for all $i$. On the other hand, $f(x)-n^e$ has an integral root for each $n$. Thus there is an index $i$ such that $P_i(t,x)$ has degree $1$ with respect to $x$, so $f(x)-t^e$ has a root in $\mathbb Q(t)$. Gauss' lemma yields that this root is in $\mathbb Q[t]$, so $f(t)=g(t)^e$ for a polynomial $g$ with rational coefficients. One quickly gets that the coefficients of $g$ then are indeed integers.

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Yes. See e.g. http://www.isibang.ac.in/~sury/polys.pdf Corollary 3.3.

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