, 2015

Suppose for the sake of contradiction that
$$P(X)=Q(X) H(X) R(X),$$
where $Q(X)$, $H(X)$, $R(X)$ are non-constant polynomials with integer coefficients. Since $P(x)>0$ for every $x \in \mathbb{R}$, the degrees of $Q(X)$, $H(X)$, $R(X)$ are all even. It implies that two of these three polynomials are quadratic. Suppose that $\operatorname{deg} Q(X)=\operatorname{deg} H(X)=2$.
Now, $P(1)=P(11)=23$, implies that $Q(1)$, $Q(11)$ are divisors of $23$. This means that $Q(1)$, $Q(11) \in\{ \pm 1, \pm 23\}$. But because $10$ divides $Q(11)-Q(1)$ we have $Q(1)=Q(11)$. Similarly, we have $H(1)=H(11)$.
Besides, $Q(1) H(1)$ is a divisor of $23$ so at least one of $Q(1)$ or $H(1)$ is $\pm 1$. Suppose without loss of generality that $Q(1)= \pm 1$ then $Q(11)=Q(1)= \pm 1$. This implies that $Q(X)=(X-1)(X-11) \pm 1$. But this implies that $Q(X)$ has a real root while $P(x)$ is positive for all $x \in \mathbb{R}$, which is a contradiction.