Polynomials, derivatives and repeated roots 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.
 A: 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. 
A: 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.)
A: This is assuming you are talking about polynomials over $\mathbb{C}$ or $\mathbb{R}$.
Assume $f$ has leading coefficient $1$ and is degree $n$.
$nf(x) = f'(x)(x-a)$
Thus
$f'(x)/nf(x) = 1/(x-a)$
Integrate
$\log(f(x)) = n\log(x-a) + C$
Thus $f(x) = K(x-a)^n$
A: If $\,n = \deg P = 0\,$ then $\,P \equiv 0\,$, else if $\,n \ge 1\,$ then $\,P(x)=\lambda(x-a)P'(x)\,$ for some $\,\lambda, a \in \mathbb Q\,$.
Suppose $P'$ had a (possibly complex) root $b \ne a$ of multiplicity $k \ge 1$, then the LHS would have $b$ as a root with multiplicity $k+1$ while the RHS would have it with multiplicity $k \lt k + 1$ which contradicts the equality of the two sides. Therefore, any roots of $\,P'\,$ must be equal to $\,a\,$, so $\,P(x) = \mu (x-a)^n\,$.
Let $\,a = p/q\,$ with $\,p, q \in \mathbb Z\,$, $\,\gcd(p,q)=1\,$. Since $\,P(x) = \mu (x-a)^n \in \mathbb Z[x]\,$ it follows that $\,\mu \in \mathbb Z\,$ and $\,\mu a^n = \mu p^n / q^n \in \mathbb Z\,$, thus $\,q^n \mid \mu\,$ so $\,\mu = cq^n\,$ for some $\,c \in \mathbb Z\,$, and in the end $\,P(x) = c(q x - p)^n\,$ with $\,c, q, p \in \mathbb Z\,$ where $\,c, q \ne 0\,$ for the degree to be $\,n\,$.
