Let a is any positive even integer

Since we know by Euclid algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying

a = bq + r where 0 <= r < b

Here b = 4, then

a = 4q + r where 0 <= r < 4

Since 0 <= r < 4, then possible remainder are 0, 1, 2 and 3.

Now possible values of a can be 4q, 4q + 1, 4q + 2, 4q + 3

Since a is even, a cannot be 4q+1 or 4q + 3 as they are both not divisible by 2.

Hense,  any even integer is of the form 4q or 4q + 2.