The shortest distance from the circle $x^2 + y^2 = 4x + 8y$ to the point $(5,-2)$ can be written in the form $\sqrt{m}$, where $m$ is an integer. Find $m$.
Completing the square gives $(x-2)^2 + (y-4)^2 = 20$, so the circle has radius $\sqrt{20} = 2\sqrt{5}$ and center $(2,4)$. The distance between $(2,4)$ and $(5,-2)$ is given by $\sqrt{(2-5)^2 + (4-(-2))^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$. Hence, the shortest distance is the difference of the distance between the center and the point and the radius, yielding $3\sqrt{5} - 2\sqrt{5} = \sqrt{5}$. Thus, $m = \boxed{5}$.

[asy]
import graph; size(8.33cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.5,xmax=8.83,ymin=-4.5,ymax=9.58;

pen ttzzqq=rgb(0.2,0.6,0);
Label laxis; laxis.p=fontsize(10);

xaxis(-3.5,8.83,defaultpen+black,Ticks(laxis,Step=2.0,Size=2),Arrows(6),above=true); yaxis(-4.5,9.58,defaultpen+black,Ticks(laxis,Step=2.0,Size=2),Arrows(6),above=true); draw(circle((2,4),4.47)); draw((2,4)--(5,-2)); draw((4,0)--(5,-2),linewidth(1.6)+ttzzqq);

label("$(x - 2)^2 + (y - 4)^2 = 20$",(0.91,5.41),NE*lsf); dot((5,-2),ds); label("$(5, -2)$",(5.15,-1.75),NE*lsf); dot((2,4),ds); dot((4,0),ds);

clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);

[/asy]