Choose a prime number greater than 3.
Multiply it by itself and add 14.
Divide by 12 and write down the remainder.
It will always be 3.
Any prime greater than 3 has to be of the form 6n+1 or 6n-1
(6n+1)^2 = 36n^2 + 12n + 1
(6n-1)^2 = 36n^2 - 12n + 1
adding 14 makes 36n^2 + 12n + 15 or 36n^2 - 12n +15
In either case first two terms have 12 as a factor and so give remainder zero with 12. Hence net remainder is same as remainder of 15 with 12 which is 3.
The problem can be modified
Multiply it by itself and add 14
14 can be changed to any number n, remainder when divided by 12 will be (n+1)Mod12.
Same rule can be obtained when dividing and checking remainder with 6, 4, 3, 2 also.