The actual contest ask to prove that the number of complex roots with absolute value less than $1$ is odd. Which can be done by adding some more easy work. Anyways, here is my solution during exam. Let $A=P(1), B=P(-1)$. Then the problem's condition is equivalent to $|A-B|>|A+B|\implies (A-B)^2>(A+B)^2\implies A^2-2AB+B^2>A^2+2AB+B^2\implies AB<0$ Now suppose that there're exist odd number of real roots with absolute value less than $1$ namely, $r_1,r_2,...,r_{2k}$ (multiple roots appear many times). Let $Q(x)=(x-r_1)(x-r_2)...(x-r_{2k})$. Therefore $Q(1)Q(-1)=(1-{r_1}^2)(1-{r_2}^2)...(1-{r_{2k}}^2)>0$ since each bracket is negative. Let $R(x)=\dfrac{P(x)}{Q(x)}$ therefore $R(1)R(-1)=\dfrac{P(1)P(-1)}{Q(1)Q(-1)}<0$. By IVT $R(x)$ has at least $1$ real root with absolute value less than $1$ which is impossible since we have divided out all such root.