Problem - 3939
Let integers $u$ and $v$ be two integral roots to the quadratic equation $x^2 + bx+c=0$ where $b+c=298$. If $u < v$, find the smallest possible value of $v-u$.
\begin{align*}
(v-u)^2 &= (v+u)^2-4uv\\
&= b^2 + 4c \\
&= b^2 + 4\cdot(298-b)\\
&= b^2 + 4b -4\cdot 298\\
&= (b+2)^2 - 1196
\end{align*}
Therefore $(b+2)\ge \sqrt{1196}\approx 35$. When $(b+2)=35$, there is no integer solution for $(v-u)$. But when $(b+2)=36$, we find $(u-v)=10$ satisfy the condition. Therefore the answer is $\boxed{10}$.