TITLE: Can conjugate real algebraic numbers always be distinguished by the sign of an iterated derivative of their minimal polynomial?
QUESTION [6 upvotes]: Let $P$ be an irreducible polynomial with rational coefficients and degree $d$. Let $\alpha$ and $\beta$ be two real roots of $P$. Then the numbers $a_k=P^{(k)}(\alpha)$
and $b_k=P^{(k)}(\beta) \ (1\leq k\leq d-1)$ are real numbers.
If $a_kb_k \gt 0$  for $1\leq k\leq d-1$, does it
always follow that $\alpha=\beta$ ?

REPLY [4 votes]: Yes. This has nothing to do with algebraic numbers; the following statement is true: 

Let $P(x)$ be a real polynomial. Suppose that $P(\alpha) = P(\beta)$
  with $\alpha$ and $\beta$ real numbers, $\alpha<\beta$. Then there is
  some $k$ for which $P^{(k)}(\alpha) P^{(k)}(\beta) < 0$.

Note that $P^{(\deg P)}(x)$ is a nonzero constant and, in particular, is not $0$ on the interval $(\alpha, \beta)$. So there exist positive integers $n$ for which $P^{(n)}(x)$ is nonzero on $(\alpha, \beta)$. Let $m$ be the least such positive integer. By Rolle's theorem, $m>1$.
Since $\frac{d}{dx} P^{(m-1)}(x)=P^{(m)}(x)$, we see that $P^{(m-1)}(x)$ is monotone on $(\alpha, \beta)$. Since $m$ was chosen minimal, $P^{(m-1)}(x)$ has a zero in $(\alpha, \beta)$. (Here we use the observation that $m>1$, so $m-1$ is a positive integer.) So $P^{(m-1)}(\alpha)$ and $P^{(m-1)}(\beta)$ have different signs. 
This would have made a nice Putnam problem.