A divisibility problem

Prove that any number which is NOT divisible by 2 and 3. Then square of the number -1 will be divisible by 24
This innocent looking problem is indeed quite easy. First thing is to observe is that the number is odd because it's not divisible by 2. Let us assume that the number is x and according to the given conditions 2 and 3 do not divide x^2-1. Which means. Now x^2-1 can also be written as (x-1)(x+1) and since x is odd it means both x-1 and x+1 are even and since both of these happen to be consecutive even that means one of them has to be a factor of 4 and obviously the other is a factor of 2 and together their product has a factor 8. Since we have to prove that 24 is the factor and so far we already proved that 8 is the factor. How to prove that 3 is also one of the factor. Well it's pretty simple too both the number x-1 and x+1 are even and according to the problem x is not divinely by 3 therefore the other two remainders are 1 and 2. Suppose x is 1 bigger than x-1 will be the factor and if x is 2 bigger than x-2 will be the factor. Thus we have a guaranteed 3 as a factor between the product x-1 and x-2 when x is not divisible by 3. Hence the whole thing is divisible by 24.