The squares of the numbers at either side of multiples of 3 progressively move up in multiples of 12 e.g. 4^2 - 2^2 = 12 & 7^2 - 5^2 = 24.
This can be proved by letting n = any number. Therefore 3n must equal any multiple of 3 as 3 is a factor. 3n + 1 is one digit above any multiple Of 3 and 3n - 1 one digit below.
Therefore, the difference between the squares of these values = (3n + 1)^2 - (3n-1)^2 = (9n^2 + 6n + 1) - (9n^2 - 6n +1) by expansion of brackets.
The terms of degree zero and two (9n^2 & 1) cancel out leaving 6n - - 6n = 12n which is proof that multiples of 3 (where n indicates the number of the multiple), goes up in multiples of 12.
The same proof can be applied to multiples of 5 and the squares of either side of them.
For multiples of 5, the resulting values go up in multiples of 20 because the terms of degree zero and two cancel out leaving the term of degree one (terms to the power of one), whose sum is 20 in this case.
For other multiples e.g. 6, differences between two squares at either side is 24, 7 is 28 and 8 is 32 and so on where resulting differences keep going up in multiples of 4.
This can also be proved by creation and expansion of brackets:
Let m = any multiple of a number. The differences of the squares of the numbers one digit above and one digit below this number can be represented, once again using algebra, as (m+1)^2 - (m-1)^2 = (m+2m+1) - (m-2m+1) which equals 2m - -2m = 4m.