How does aircraft weight affect landing distance?
If you discount aerodynamic effects, and assume the same landing speed, on a theoretical level mass makes no difference. The additional kinetic energy to be dissipated is exactly cancelled by the additional normal force. The kinetic energy is given by the formula:

E = 1/2 mv², where


E is the kinetic energy
m is the vehicle's mass
v is the speed at the start of braking


The work done by braking is given by:

W = μmgd, where 


W is work done
μ is the coefficient of friction between the road surface and the tires
g is the gravity of Earth
d is the distance travelled


The braking distance given an initial speed v is then found by putting W = E, from which it follows that

d = v²/2μg 


d = minimum stopping distance
v = velocity when brakes were applied
μ = coeff. of friction between wheels and road
g = gravitational constant


Mass is removed from the equation. The kinetic energy is tranferred into heat, so the larger the mass the more heat is generated. But theoretically the stopping distance is unaffected.

In reality, though, the coefficient of friction of tires does not remain constant with changing normal force. As normal force increases the coefficient of friction (μ) of tires decreases. 

Since the μ does not remain linear with mass a heavier plane will have an increase in kinetic energy which is higher than the corresponding increase in friction of the tires, thus a longer stopping distance. 

All of this assumes that “maximum braking” means you have as much brake power available as the maximum friction force of the tires. Above a certain weight/speed this will not be the case for an airliner. They are designed for a particular amount of deceleration force at MTOW and that will be less than the amount of friction tires can provide on a dry runway. (See this question for more details on that.) That will also cause a heavier plane to roll farther.