amparvardi wrote: Let $p: \mathbb C \to \mathbb C$ be a polynomial with degree $n$ and complex coefficients which satisfies \[x \in \mathbb R \iff p(x) \in \mathbb R.\] Show that $n=1$ Assume that $p$ satisfies the condition given above, and let $\deg p = n$. $p$ obviously has real coefficients, since it is determined by $p(0), p(1), \ldots, p(n)$. Assume that $n > 1$. For any $s \in \mathbb{R}$ the polynomial $p(x)-s$ must have only real roots. For $n=2$ we easily reach a contradicition, so $n \ge 3$. Let $r_1 \le \ldots \le r_{n-1}$ be the roots of $p'(x)$ (counted with root multiplicity). The roots of $p(x)-s$ must then be in the intervals $(-\infty,r_1], [r_1,r_2],\ldots,[r_{n-1},\infty)$. Then just choose $s$ s.t. $f(r_1), f(r_2)$ has the same sign, then there cannot be a real root in the interval $[r_1,r_2]$. Contradiction. Hence $n=1$.