Given a polynomial f, can there be more than one constant c such that every root of f(x)-c is repeated? The question
Let $f$ be a nonconstant polynomial over $\mathbb{C}$.  Let's say that a point $c \in \mathbb{C}$ is unusual for $f$ if every root $x$ of $f(x) - c$ is repeated.  Can $f$ have more than one unusual point?
Short remarks


*

*There can be exactly one unusual point, e.g. if $f(x) = x^2$.  There can be none, e.g. if $f(x) = x^3 + 3x$.

*There are at most $\deg(f) - 1$ unusual points, since every unusual point is the image under $f$ of a critical point. 

*The hypotheses "nonconstant" and "over $\mathbb{C}$" could be varied.  I added them to rule out cases like the following: over a field $k$ of characteristic $p$, every point of $k$ is unusual for $x^p$ (since the derivative of $x^p$ is $0$).  I'd be happy to change the hypotheses to "nonconstant polynomial over an algebraically closed field of characteristic 0".  Maybe something like "polynomial whose derivative is nonzero, over an algebraically closed field" would also be sensible.
Weak reason to expect the answer to be "no"
Perhaps there's a very short answer to my question: I could be overlooking something elementary.  But in case it's not so easy, I'll give a flimsy argument for why we might expect the answer to be "no" - that is, for why we might expect every polynomial to have at most one unusual point.  
My question is equivalent to: is there a nonconstant polynomial $f$ over $\mathbb{C}$ for which $1$ and $-1$ are both unusual?  For $1$ to be unusual means that every root of $f(x) - 1$ is also a root of $f'(x)$.  Writing $d = \deg(f)$, this is equivalent to 
$$
(f(x) - 1) \mid f'(x)^d.
$$
Similarly, for $-1$ to be unusual means that $(f(x) + 1) \mid f'(x)^d$.  Both together are equivalent to 
$$
(f(x)^2 - 1) \mid f'(x)^d,
$$
that is,
$$
(f(x)^2 - 1)\cdot g(x) = f'(x)^d
$$
for some $g(x) \in \mathbb{C}[x]$.  This forces $\deg(g) = d(d - 3)$ (and so $d \geq 3$).  
So, can we find $f$ and $g$ satisfying the last displayed equation?  Comparing coefficients, what we have here is a system of $d(d - 1) = d^2 - d$ equations in $(\deg(f) + 1) + (\deg(g) + 1) =  d^2 - 2d + 2$ unknowns.  There are $d - 2$ more equations than unknowns, and $d \geq 3$, so a first guess is that it can't be done. 
 A: The question can be restated as follows: can there be a branched $n$-fold cover $S^2\to S^2$ with at least 3 critical values, one of index $n$ and all preimages of critical values being critical points?
If such a map existed, then in the preimage of any critical point other than the infinity we would have $\leq \frac{n}{2}$ points. The Riemann-Hurwitz formula for a branched $n$-fold cover $M\to N$ of 2-manifolds reads $$\chi(M)=n(\chi(N)-k)+a_1+\cdots +a_k$$ where $k$ is the number of the critical points and $a_i$ is the cardinality of the preimage of the $i$-th critical point. In our case this gives $$2=n(2-k)+1+a$$ with $a\leq\frac{n}{2}(k-1)$, so the right hand side is $\leq n(2-k+\frac{k-1}{2})+1=n(\frac{3}{2}-\frac{k}{2})+1$ and can't be 2 when $k\geq 3$.
A: Noam Elkies left a very useful comment underneath Qiaochu Yuan's answer, providing a completely elementary and self-contained solution to my question.  For the benefit of anyone interested, I'll spell it out here.  (And I'll make this answer community wiki as this is Noam's solution, not mine.)
Let $k$ be an algebraically closed field.  For $f \in k[x]$ and $\alpha \in k$, write $\mu(f, \alpha)$ for the multiplicity of $\alpha$ as a root of $f$; that is, $\mu(f,\alpha) = \sup \{n \in \mathbb{N} : (x - \alpha)^n|f(x)\}$.  Some very basic facts:


*

*$\mu(f,\alpha) > 0$ iff $f(\alpha) = 0$

*$\sum_{\alpha \in k} \mu(f,\alpha) = \sum_{\alpha \in f^{-1}(0)} \mu(f,\alpha) = \deg(f)$ as long as $f \neq 0$

*$\mu(f',\alpha) \geq \mu(f,\alpha) - 1$.
A point $a \in k$ is unusual for $f$ iff $\mu(f-a,\alpha) \geq 2$ for all $\alpha \in f^{-1}(a)$.  
For nonconstant $f$, it's easy to see that if $a$ is unusual then $|f^{-1}(a)| \leq \deg(f)/2$.  Indeed,
$$
\deg(f) = \deg(f - a) = \sum_{\alpha \in f^{-1}(a)} \mu(f - a,\alpha) \geq 2|f^{-1}(a)|.
$$
Theorem  Let $f \in k[x]$.  If $f' = 0$ then every point of $k$ is unusual for $f$.  If $f' \neq 0$ then at most one point of $k$ is unusual for $f$.
Proof  The first statement is clear.  For the second, suppose that $a$ and $b$ are unusual, with $a \neq b$.  We have
$$
\begin{aligned}
\sum_{\alpha \in f^{-1}(a)} \mu(f', \alpha)& = \sum_{\alpha \in f^{-1}(a)} \mu((f - a)', \alpha)\\\
&\geq \sum_{\alpha \in f^{-1}(a)} \bigl[\mu((f - a), \alpha) - 1\bigr]\\\
& = \deg(f-a) - |f^{-1}(a)|\\\
& \geq \frac{1}{2}\deg(f).
\end{aligned}
$$
The same goes for $b$.  But $f^{-1}(a) \cap f^{-1}(b) = \emptyset$ and $f' \neq 0$, so
$$
\begin{aligned}
\deg(f') & = \sum_{\gamma \in k} \mu(f', \gamma)\\\
& \geq \sum_{\alpha\in f^{-1}(a)} \mu(f', \alpha) + \sum_{\beta\in f^{-1}(b)} \mu(f',\beta)\\\
& \geq \frac{1}{2}\deg(f) + \frac{1}{2}\deg(f) = \deg(f),
\end{aligned}
$$
giving $\deg(f') \geq \deg(f)$, a contradiction.
A: This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero). 
We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and $f, h$ have all of their roots repeated. If $g$ is nonzero, $f, h$ must be relatively prime. Letting $d = \deg f$, it follows that $fgh$ has at most $d$ roots, but by Mason-Stothers $fgh$ must have at least $d+1$ roots; contradiction. 
A: Encouraged by Pierre-Yves Gaillard here is a quick solution to the original problem. It is based on Tom Leinster's answer which in turn is an elaboration of Noam Elkies's comment to Qiaochu Yuan's answer.
Assume $a\neq b$ are unusual for $f(x)$. Then $f'(x)^2$ is divisible by $f(x)-a$ and $f(x)-b$, hence also by their product. This would show $\deg(f')\geq\deg(f)$, a contradiction.
A: Extend $f$ to a regular morphism $\bar f:\mathbb P^1\to \mathbb P^1$ and write down the Hurwitz formula: 
$$K_{\mathbb P^1}\sim \bar f^*K_{\mathbb P^1} +R$$
where $R$ is the ramification divisor. Since $f$ is a polynomial, $\bar f$ is completely ramified at $\infty$, so $R$ contains that point with multiplicity $d-1$. Therefore, by the above equivalence the rest of $R$ has degree $d-1$. 
Now for a point on the target (your "$c$") for which all the points in the preimage are multiple, the degree of the ramification divisor above this point has to be at least $d-\frac d2=\frac d2$ (it's the degree of the map minus the number of points). If you had two such points their combined degree would be at least $d$ contradicting the previous observation that it should be at most $d-1$. 
