The statement of nworth that the exposure can be scaled as the square of the lens distance is correct. (To be really precise, the image distance).
Some of the other statements are either imprecise or not accurate.
If you want to figure out the exposure time compensation by measuring linear dimensions of the prints, i.e., by the two magnifications, the correct formula is (M2 + 1) ^ 2 / (M1 + 1)^2. That is, the square of the ratios of the factors (magnification plus one). It seems obvious that the inverse square law should use the just the magnifications, but this neglects refocusing the lens and isn't accurate. The refocusing changes the effective f-stop.
For a derviation, see Lenses in Photography by Rudolf Kingslake or Applied Photographic Optics by Sidney Ray.
For example, going from M1 = 2 (8x10 from a 4x5 neg) to M2 = 4 (16 x 20 from a 4x5 neg), the change in exposure time should be x2.78 longer, not x4 longer. If you are doing enlargements from a small negative, the "+1" tends not to matter, for example, an 8x11 from 35 mm is M=8 and M and M+1 are about the same, and simplifying the equation gives an answer that is close.
Of course, reciprocity effects or artistic choices depending on print size might cause departures from this equation, which only accounts for light.
It also works to square the ratio of image distances because d_image = focal length * (magnifcation + 1). Maybe this is another way to see that there should be a "+1" in the magnification version of the equation.
For the numeric example above, the 8x10 print from 4x5 neg has d_image = 450 mm and the 16x20 has d_image = 750 mm. The ratio squared is 2.78.
In this example, the two distances between head and the paper (i.e., the the negatives and paper) are 675 mm and 937.5 mm. It doesn't work to take the ratio these distances and square.