Problem (Combinatorics):
Suppose that $u$ and $v$ are distinct numbers chosen at random from the set $\{1,2,3, \ldots, 30\}$. Compute the probability that the roots of the polynomial $(x+u)(x+v)+4$ are integers.

Solution:
Assume without loss of generality that $u>v$. The condition that $(x+u)(x+v)+4$ has integer roots is equivalent to the discriminant $(u+v)^{2}-4(u v+4)=(u-v)^{2}-16$ being a perfect square. This is possible if and only if $u-v=4$ or $u-v=5$. There are $(30-4)+(30-5)=26+25=51$ such ordered pairs $(u, v)$, so the answer is

$$
\frac{51}{\left(\begin{array}{c}
30 \\
2
\end{array}\right)}=\frac{\mathbf{1 7}}{\mathbf{1 4 5}}
$$

Final answer: $\frac{17}{145}$