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Given a polynomial $f \in \mathbb{Q}[x]$ such that $f^2 \in \mathbb{Z}[x]$, is $f \in \mathbb{Z}[x]$?

In words, if the square of a rational polynomial has integer coefficients does it have integer coefficients as well?

I doubt this is true, but it appears to be true for $f$ with degree less than 3 (by inspection of the coefficients). For $\text{deg}f \ge 3$ I can't tell based on looking at coefficients, and I haven't found a counterexample yet.

  • I think if the polynomial can be written as a square then yes, because if $f$ has integer coefficients then $f^2$ does as well, you can test this by squaring the general polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ where $a_n,a_{n-1}...a_0$ are all integers – Pen and Paper Oct 01 '21 at 06:11
  • Interesting question! I wish I had an answer. Curiously, my gut tells me it IS true, though I have nothing concrete to base that on. – A.J. Oct 01 '21 at 06:12
  • @Quippy That's not sufficient to show $f$ is integral. If $f$ has integer coefficients it's clear that $f^2$ does, but the converse is not clear (and is the purpose of this question). If you square your general polynomial you'll find the coefficients of $f^2$ are quadratic combinations of the coefficients of $f$. These combinations may be integral without the coefficients of $f$ being integral (as far as I can tell). –  Oct 01 '21 at 06:24
  • $f$ is an integer-valued polynomial; does it help to express it as a $\mathbb{Z}$-linear combination of the $x\choose k$'s? – ancient mathematician Oct 01 '21 at 06:40
  • @ancientmathematician I don't see why $f$ is integer-valued here. $f^2$ is certainly integer-valued though. –  Oct 01 '21 at 06:47
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    $f$ is integer valued because $f(n)= a/b$ is rational for every $n$, and $f^2 = a^2/b^2$ is an integer. That means $b^2$ divides $a^2$ and therefore $b$ divides $a$, so $f(n)$ was already an integer. – A. Thomas Yerger Oct 01 '21 at 07:04
  • ah, makes sense @AlfredYerger. That is more than I had, but not enough to show $f$ is (or isn't) integral as far as I can tell. –  Oct 01 '21 at 07:23
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    This should be a consequence of Gauss's lemma, that if you have $F(x) \in \mathbb{Z}[x]$ and it factors into $F(x)=G(x)H(x)$ with $G(x), H(x) \in \mathbb{Q}[x]$ then really $G(x), H(x) \in \mathbb{Z}[x]$. In our case, $F(x)=f(x)^2$ and $G(x)=H(x)=f(x)$. – Merosity Oct 01 '21 at 10:07
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    This is also a special case of $p(q(x))$ cannot have only integer cofficients if $q(x)$ has some non-integer coefficient (see https://math.stackexchange.com/questions/3793239), where $p(x)=x^2,q(x)=f(x)$ in your case (i had the additional condition of $q$ being monic, but answers show it's not needed) – Sil Oct 01 '21 at 17:10
  • Obviously not. Counterexample: lt p = a + b x, a=b=sqrt(2), then p^2 = 2 + 4x + 2x^2 has integer coefficients but p has not. – Dr. Wolfgang Hintze Oct 02 '21 at 14:40
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    @Dr.WolfgangHintze $\sqrt{2}+\sqrt{2}x$ does not have all coefficients rational (the $f \in \mathbb{Q}[x]$ assumption) – Sil Oct 02 '21 at 16:32

2 Answers2

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Yes, the original polynomial has integer coefficients.
Take any prime $p$ in a denominator of a term. Let the coefficients be $$a_i=p^{r_i}b_i$$ where the $b_i$ contain no factors of $p$, in either their numerator or denominator.
The values $r_i$ may be positive or negative, but by assumption at least one is negative. Let $$-R=\min_i r_i\\k=\min\{i|r_i=-R\}$$
Now calculate $(f(x))^2$. In the sum of terms that contributes to $x^{2k}$, the only term with $p^{2R}$ in the denominator is $(a_k)^2$. No factor has larger than $p^{R}$ .in the denominator , and terms to the right of $x^k$ have smaller powers of $p$.
So, over a common denominator, that is the only term not a multiple of $p$. So the full coefficient of $x^{2k}$ will not have a multiple of $p$ in the numerator. It won't cancel, and that coefficient won't be an integer.

Empy2
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That’s just a reformulation of arguments made in the comments and the previous answer, but it might be clearer: there exists a minimal positive integer $N$ such that $f_1:=Nf$ has integer coefficients. Then $f_1^2=N^2f^2$ has coefficients divisible by $N^2$.

So if $p$ is a prime dividing $N$ and $g_1$ is the reduction mod $p$ of $f_1$, then $g_1^2=0$. Thus $g_1=0$, and therefore $f_1/p=N/p \times f$ has integer coefficients, contradicting the minimality of $N$. So $N=1$ and we are done.

Aphelli
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