Why can’t a non-constant polynomial have a constant interval? Given a polynomial that is not constant, of course, it doesn’t contain a constant interval. But how can we prove it?
Since the polynomial is differentiable over R, I came up with a solution that uses Lagrange mean value theorem n times and reduces the nth derivative to a constant. Since the leading term is not zero, there can not be a zero in the nth derivative, and that contradicts the mean value theorem. Therefore, the polynomial must not contain a constant interval.
However, I do realize that this solution is a bit complicated,  so is there a simpler solution(possibly elementary) that can prove this?
 A: A polynomial of degree $n \gt 0$ has at most $n$ roots. A constant polynomial is of degree $0$.
A non constant polynomial $p$… can’t be constant. Therefore it is of degree $n\gt 0$. If it would be constant and equal to $a$ on an interval of strictly positive lenght, $p-a$ would have an infinite number of roots. A contradiction.
A: Suppose $p$ is a polynomial of degree $n$, i.e. $$ p(x) = \sum\limits_{i=0}^n a_ix^i $$
and $p(x) = c$ for $x\in (a,b)$. Choose $n+1$ distinct points from $(a,b)$, $x_0,\dots , x_n$. The system of equations $p(x_i) = c$ for the $a_i$ is clearly solved by $a_0 = c$ and $a_i = 0$ otherwise. However, since the matrix of coefficients (the Vandermonde matrix) is invertible whenever the $x_i$ are distinct*, the solution is unique. Thus, $p(x) \equiv c$.
A: Suppose $P(x) = c$ for all $x \in [a, b]$ with $a < b$ and $P$ is nonconstant. Then let $n > 0$ be the degree of the polynomial. Then $Q(x) = P(x) - c$ is also of degree $n$. But $Q$ has roots $a + \frac{i}{n} (b - a)$ for $0 \leq i \leq n$, which is $n + 1$ roots. And a polynomial of degree $n$ with coefficients in a field can have at most $n$ roots. Contradiction.
