It can be proved that the squares of prime numbers greater than 3 are all multiples of 24 plus 1.

Because it is a prime number, it can be expressed as 6k+ 1 or 6k- 1.

(6k+ 1)^2=36k^2+ 12k+ 1= 12k(3k+ 1)+ 1

So whether k is odd or even, k and 3k+ 1 must be odd and even.

So the square of 6k+ 1 minus 1 is a multiple of 24.

Similarly, (6k-1) 2 = 36k2-12k+1=12k (3k-1)+1.

K and 3k- 1 must be odd and even.

So the square of 6k- 1 minus 1 is a multiple of 24.

So the square of prime number-1 must be a multiple of 24.