, 2016

Solution:

The construction for $n=9$ can be achieved with the polynomials $x^{2}+x+1$, $x^{2}+x+2$, and $x^{2}+5$.

First we consider what kinds of polynomials we can have. Let $p(x) = (x+h)^{2} + k$. $h$ is either an integer or half an integer. Let $k=0$. If $h$ is an integer then $p(x)$ hits the perfect squares $0, 1, 4, 9$, etc. If $h$ is half an integer, then let $k=1/4$. Then $p(x)$ hits the product of two consecutive integers, i.e. $0, 2, 6, 12$, etc.

Assume there is a construction for $n=10$. In both of the cases above, the most a polynomial can hit out of $10$ is $4$, in the $0, 1, 4, 9$ case. Thus $p_{1}$ must hit $1, 2, 5, 10$, and $p_{2}$ and $p_{3}$ hit $3$ integers each, out of $3, 4, 6, 7, 8, 9$. The only ways we can hit $3$ out of $7$ consecutive integers is with the sequences $0, 2, 6$ or $0, 1, 4$. The only way a $0, 2, 6$ works is if it hits $3, 5$, and $9$, which doesn't work since $5$ was hit by $p_{2}$. Otherwise, $p_{2}$ is $0, 1, 4$, which doesn't work as $p_{2}$ hits $3, 4$, and $7$, and $p_{3}$ must hit $6, 8$, and $9$, which is impossible. Thus no construction for $n=10$ exists.