As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude? From Earth, Jupiter's maximum brightness is -2.94 and Saturn's is -0.24. But what about from Mars? They should be brighter, but by how much?

There are equations at the obvious wiki entry but I'm not sure I understand them. Plus I'm afraid to attempt this calculation myself because I've heard that this sort of thing does not follow the inverse square law. They're reflecting light (not generating it), and I read somewhere that it should follow an inverse 4th power law because of that. I don't see any 4th powers in the apparent magnitude equations.
The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected non-specularly  — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the light has to go back and forth. But if the reflector scatters light in all directions — i.e. into a $2\pi$ hemisphere — then the detected flux is $\sim r^2/d^4$, where $r$ is the radius of the reflector (see this answer for a more throrough explanation).

An example of this is a radar. But in our case, it isn't us that emit the light, it's the Sun. The amount of light reflected from Jupiter and Saturn depends on their distance to the Sun, and that distance doesn't change if you move to Mars. The relevant distances (that I got from NASA's Planetary Fact Sheet) are:


Earth semi-major axis    $d_\mathrm{E} = 1.00\,\mathrm{AU}$
Mars aphelion$^\dagger$  $d_\mathrm{M} = 1.64\,\mathrm{AU}$
Jupiter semi-major axis  $d_\mathrm{J} = 5.20\,\mathrm{AU}$
Saturn semi-major axis   $d_\mathrm{S} = 9.58\,\mathrm{AU}$


Now the differences between them:


Earth to Mars    $d_\mathrm{M-E} = 0.64\,\mathrm{AU}$
Earth to Jupiter $d_\mathrm{J-E} = 4.20 \,\mathrm{AU}$
Earth to Saturn  $d_\mathrm{S-E} = 8.58\,\mathrm{AU}$
Mars  to Jupiter $d_\mathrm{J-M} = 3.56 \,\mathrm{AU}$
Mars  to Saturn  $d_\mathrm{S-M} = 7.94 \,\mathrm{AU}$


Hence, from Mars the distance to Jupiter is $d_\mathrm{M-J} = 0.85 d_\mathrm{J-E}$, and the received flux  is thus $1/0.85^2 = 1.4$ times that on Earth. The change in apparent magnitude is then
$$
\Delta m = -2.5 \log\left( \frac{0.85^2}{1}\right) = -0.36,
$$
i.e. Jupiter would be $m = -2.94 - 0.36 = -3.30$ as viewed from Mars (assuming the values you provide are correct; I didn't check this).

Following the same approach for Saturn, I get $m = -0.41$.

$^\dagger$The reason I used Mars' aphelion instead of Mars' semi-major axis is because Mars has a rather eccentric orbit ($e\sim0.09$). Jupiter and Saturn are somewhat closer to circular orbits, though ($e\sim0.05$). This of course is still an approximation. If you wish to take into account the eccentricities of all orbits, you also need to know the angle between their semi-major axes. This also doesn't take into account the inclination of the orbit; however, these are only 1º-2º.
And of course this minimum distance will not occur every Martian year.