A cute problem from Turkish AYT exam

 $P(x)$ is a $4^{th}$ degree polynomial with real coefficients that satisfies

$$P(x) \ge x \quad \forall x \in R$$

$$P(1) = 1, P(2) = 4, P(3) = 3$$

Find the value of $P(4)$.

Scroll down for a solution.

Let $H(x) = P(x) - x$ and so $H(x) \ge 0 \forall x \in R$. Since $H(1) = H(3) = 0$, $H(x)$ has at least two distinct roots.

Now if there was a root of $H$ different from $1$ or $3$, then we can show that $H(c) < 0$ for some $c \in R$. If the multiplicity of $1$ of $3$ was odd, then we can again show that $H(c) < 0$ for some $c$.

Thus we must have that

$$H(x) = A(x-1)^2(x-3)^2, A > 0$$

Since $H(2) = P(2) - 2 = 2$, we get  $A = 2$.

This gives $P(4) = H(4) + 4 = 2.3^2.1^2 + 4 = 22$.