I want to describe the polynomials with integer coefficients and the property that $f'(x) \mid f(x)$ (the derivative divides the polynomial).

So I know that $f(x)$ divides $g(x)$ if all of $f(x)$'s roots are roots of $g(x)$. I also know that $f'(x)$ has degree one less than $f(x)$. Finally, $f'(x)$ detects multiple roots. If $a$ is a root of $f(x)$ and $f'(x)$, it is at least a double root.

Using these facts I've conjectured that $f(x)$, if it has degree $n$, has 1 root with multiplicity $n$ or 2 roots, one with multiplicity $n-1$. But I can't prove it and I fear that I am making a mistake thinking just about linear terms.

  • 1
    $\begingroup$ @Abelsh: are you asking that $f'(x)$ divide $f(x)$ in $\mathbb{Z}[x]$, or in $\mathbb{Q}[x]$? Because, for example, $(x-1)^2$ has the latter property, but not the former. $\endgroup$ – Arturo Magidin Nov 15 '10 at 15:04
  • $\begingroup$ Sorry: over the integers. $\endgroup$ – Abelsh Nov 15 '10 at 15:08
  • $\begingroup$ @Abelsh: And are the roots to be found in $\mathbb{C}$? $\endgroup$ – Arturo Magidin Nov 15 '10 at 15:09
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    $\begingroup$ @Abelsh: If you want the divisibility over $\mathbb{Z}[x]$, then the answer is that the only polynomials with the desired property are $f(x)=0$ and the linear polynomials. Otherwise, just look at the leading term: the leading term of $f(x)$ is $a_nx^n$ for some $a_n\neq 0$; the leading term of $f'(x)$ is $na_nx^{n-1}$. If $f(x)=g(x)f'(x)$ with $g(x),f'(x)\in\mathbb{Z}[x]$, then the leading coefficient $na_n$ of $f'(x)$ has to divide the leading coefficient of $f(x)$, $a_n$, and that can only happen if $n=1$. If $n=0$, then $f(x)=k$ and $f'(x)=0$, and $0$ only divides $k$. $\endgroup$ – Arturo Magidin Nov 15 '10 at 15:17
  • $\begingroup$ @Arturo: some linear polynomials. Consider $2x+3$, for example :) $\endgroup$ – Mariano Suárez-Álvarez Nov 15 '10 at 15:19

I'll assume you want the divisibility to be in $\mathbb{Q}[x]$. Write $\displaystyle f(x) = \prod_{i=1}^{k} (x - a_i)^{m_i}$ where the $a_i$ are distinct (in $\mathbb{C}$).

Lemma: $\displaystyle \frac{f'(x)}{f(x)} = \sum_{i=1}^{k} \frac{m_i}{x - a_i}$.

This is a direct consequence of the product rule and you do not need to use any calculus to prove it. (That is, you don't need to take logarithms. The derivative can be defined as a totally formal operation on polynomials and this identity holds in $\mathbb{Q}(x)$.)

If $f'(x)$ divides $f(x)$, then $\frac{f'(x)}{f(x)} = \frac{n}{x - a}$ for some $a$. Now, the functions $\frac{1}{x - a}$ are linearly independent (you also do not need to use any calculus to prove this) in $\mathbb{C}(x)$. It follows that this is possible if and only if $f$ has $a$ as a repeated root of multiplicity $n$, so $a$ is the only root.

It follows that $f$ has the form $c(px - q)^n$ for some integers $c, p, q$ with $a, p \neq 0$. (I'm ignoring the degenerate cases.)

  • $\begingroup$ You can do next the case over $\mathbb Z$: if $f'$ divides $f$ over $\mathbb Z$ then it also does over $\mathbb Q$... $\endgroup$ – Mariano Suárez-Álvarez Nov 15 '10 at 15:13
  • $\begingroup$ you mean $a$ is the only root of $f$, right? And from this you get that this can never happen if we want divisibility in $\mathbb{Z}[x]$, because the leading terms will never work out. $\endgroup$ – Arturo Magidin Nov 15 '10 at 15:14

Over $\mathbb{R}$ you can use Rolle's theorem. Between two zeroes of function lies a zero of its derivative. If f has two distinct zeroes, there is a zero of f' which is not a zero of f.

  • $\begingroup$ Unless f is constant on an interval, hence constant. $\endgroup$ – Max Nov 15 '10 at 16:03

This is assuming you are talking about polynomials over $\mathbb{C}$ or $\mathbb{R}$.

Assume $f$ has leading coefficient $1$ and is degree $n$.

$nf(x) = f'(x)(x-a)$


$f'(x)/nf(x) = 1/(x-a)$


$\log(f(x)) = n\log(x-a) + C$

Thus $f(x) = K(x-a)^n$

  • $\begingroup$ Then $f'(x)$ does not divide $f(x)$ in $\mathbb{Z}[x]$ if $n\gt 1$... (think about the leading coefficients). Of course, Abelsh has not specified yet where he wants the divisibility to hold. $\endgroup$ – Arturo Magidin Nov 15 '10 at 15:06
  • $\begingroup$ @Qia: Why can't we assume that in $\mathbb{R}[x]$? $\endgroup$ – Aryabhata Nov 15 '10 at 15:13
  • $\begingroup$ @Arturo: Given the way the question was phrased, I am confident this is $\mathbb{R}[x]$ or $\mathbb{C}[x]$. (Looks more like a high-school/beginning college level question to me). $\endgroup$ – Aryabhata Nov 15 '10 at 15:14
  • $\begingroup$ Never mind. You were also assuming that the divisibility was in a polynomial ring over a field. But you don't need the divisibility to work over R[x] or C[x], since the quotient, if it exists, must have rational coefficients. $\endgroup$ – Qiaochu Yuan Nov 15 '10 at 15:16

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