For ~60\%~ of the score an easy brute force solution was available, trying every ~M~ between ~0~ and ~10\,000~ and checking whether all given integers give the same remainder when divided by ~M~.

For the full score, we need the following deduction: if some two numbers ~A~ and ~B~ give the same remainder when divided by ~C~, then ~C~ divides ~A-B~. So for an integer ~M~ to be part of the output, it must divide the differences between each pair of input numbers. The largest such ~M~ is the greatest common divisor of the differences between input numbers. The remaining values of ~M~ are all divisors of the largest ~M~ (except for ~1~).