Italian Mathematical Olympiad

Solution:

The answer is $36$. The polynomial $p(x)+q(x)$ vanishes at $x=-1$, $x=-2$ and $x=-3$; moreover it has degree less than or equal to $3$ and has integer coefficients. Therefore, by Ruffini's theorem, it factors as $p(x)+q(x)=k(x+1)(x+2)(x+3)$ with $k$ an integer. Similarly we observe that $p(x)-q(x)=h(x-1)(x-2)(x-3)$ with $h$ an integer.
We have that
$$p(x)=\frac{1}{2}(k(x+1)(x+2)(x+3)+h(x-1)(x-2)(x-3))$$
from this equality it is easy to see that $p(x)$ has integer coefficients if and only if $k$ and $h$ have the same parity. The condition for $q(x)$ to have integer coefficients is exactly the same. The problem moreover requires that $p(x)$ and $q(x)$ be distinct, which happens if and only if $h \neq 0$.
Evaluating the two initial identities at $x=0$ we obtain $p(0)+q(0)=6k$ and $p(0)-q(0)=-6h$. Squaring and adding these two equalities we find
$$(p(0)+q(0))^{2}+(p(0)-q(0))^{2}=36\left(k^{2}+h^{2}\right)$$
that is $2 \cdot\left([p(0)]^{2}+[q(0)]^{2}\right)=36\left(k^{2}+h^{2}\right)$. In light of the previous observations, the minimum of $k^{2}+h^{2}$ is obtained for $k= \pm 1$ and $h= \pm 1$. Therefore the minimum value of $[p(0)]^{2}+[q(0)]^{2}$ is $\frac{1}{2} \cdot 36 \cdot\left(( \pm 1)^{2}+( \pm 1)^{2}\right)=36$ .