Baltic Way 1992 · Problem 10

Solution: Let $p(x) = a x^4 + b x^3 + c x^2 + d x + e$ with $a \neq 0$. From (i)-(iii) we get $b = d = 0$, $a > 0$ and $e = 1$. From (iv) it follows that $p'(x) = 4 a x^3 + 2 c x$ has at least two different real roots. Since $a > 0$, we have $c < 0$ and $p'(x)$ has three roots $x = 0$, $x = \pm \sqrt{ -c / (2a) }$. The minimum points mentioned in (iv) must be $x = \pm \sqrt{ -c / (2a) }$, so $2 \sqrt{ -c / (2a) } = 2$ and $c = -2a$. Finally, by (ii) we have $p(x) = a (x^2 - 1)^2 + 1 - a \geq 0$ for all $x$, which implies $0 < a \leq 1$. It is easy to check that every such polynomial satisfies the conditions (i)-(iv).