Question:
The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\sqrt{a}/b$, where $a$ and $b$ are positive integers, and $a$ is not divisible by the square of any prime.  Find $a+b$.

Answer:
By the distance formula, we are trying to minimize $\sqrt{x^2+y^2}=\sqrt{x^2+x^4-10x^2+25}$. In general minimization problems like this require calculus, but one elementary optimization method that sometimes works is completing the square.  We have $$\sqrt{x^2+x^4-10x^2+25}=\sqrt{(x^2-9/2)^2+(25-81/4)}.$$This expression is minimized when the square equals $0$, i.e. when $x=\pm 3/\sqrt{2}.$ For this value of $x$, the distance is $$\sqrt{25-\frac{81}{4}}=\frac{\sqrt{19}}{2}.$$Hence the desired answer is $\boxed{21}$.