, 2020

Answer: No, she cannot.

Let $x_{1} < x_{2} < \ldots < x_{2n}$ be real numbers chosen by the magician. We will construct two distinct polynomials $P(x)$ and $Q(x)$, each of degree $n$, such that the member of audience will write down the same sequence for both polynomials. This will mean that the magician cannot distinguish $P$ from $Q$.

Claim. There exists a polynomial $P(x)$ of degree $n$ such that $P(x_{2i-1}) + P(x_{2i}) = 0$ for $i = 1, 2, \ldots, n$.

Proof. We want to find a polynomial $a_{n} x^{n} + \ldots + a_{1} x + a_{0}$ satisfying the following system of equations:
$$\left\{ \begin{array}{l} (x_{1}^{n} + x_{2}^{n}) a_{n} + (x_{1}^{n-1} + x_{2}^{n-1}) a_{n-1} + \ldots + 2 a_{0} = 0 \\ (x_{3}^{n} + x_{4}^{n}) a_{n} + (x_{3}^{n-1} + x_{4}^{n-1}) a_{n-1} + \ldots + 2 a_{0} = 0 \\ \cdots \\ (x_{2n-1}^{n} + x_{2n}^{n}) a_{n} + (x_{2n-1}^{n-1} + x_{2n}^{n-1}) a_{n-1} + \ldots + 2 a_{0} = 0 \end{array} \right.$$
We use the well known fact that a homogeneous system of $n$ linear equations in $n+1$ variables has a nonzero solution. (This fact can be proved using induction on $n$, via elimination of variables.) Applying this fact to the above system, we find a nonzero polynomial $P(x)$ of degree not exceeding $n$ such that its coefficients $a_{0}, \ldots, a_{n}$ satisfy this system. Therefore $P(x_{2i-1}) + P(x_{2i}) = 0$ for all $i = 1, 2, \ldots, n$. Notice that $P$ has a root on each segment $[x_{2i-1}, x_{2i}]$ by the Intermediate Value theorem, so $n$ roots in total. Since $P$ is nonzero, we get $\operatorname{deg} P = n$.

Now consider a polynomial $P(x)$ provided by the Claim, and take $Q(x) = -P(x)$. The properties of $P(x)$ yield that $P(x_{2i-1}) = Q(x_{2i})$ and $Q(x_{2i-1}) = P(x_{2i})$ for all $i = 1, 2, \ldots, n$. It is also clear that $P \neq -P = Q$ and $\operatorname{deg} Q = \operatorname{deg} P = n$.