Positive integers can be represented in one of these 6 forms:
\(k\) here is a positive integer. Primes greater than or equal to 5 cannot be of the form \(6k\), \(6k+2\), \(6k+3\) and \(6k+4\) because they are divisible by \(6\), \(2\), \(3\) and \(2\), respectively. This means primes greater than or equal to 5 can only be of the form \(6k + 1\) or \(6k + 5\).
Assume \(p\) is of the form \(6k+1\), then:
If \(k\) is odd, then \(3k^2\) is also be odd, so \([3k^2+k]\) would be even (as the sum of two odds give an even). Let \([3k^2 + k = 2h]\) where \(h\) is some integer. Then \([p^2 - 1 = (2h)*12]\), or \([p^2 - 1 = 24h]\), so \([p^2 - 1]\) is a multiple of 24.
If \(k\) is even, then \([3k^2+k]\) would also be even, so \([p^2 - 1]\) would again be a multiple of 24.
Now assume \(p\) is of the form \(6k+5\), then:
If \(k\) is odd, then \(3k^2\) and \(5k\) would also be odd, so \([3k^2 + 5k]\) would be even. If \(k\) is even, then \([3k^2 + 5k]\) would still be even. In both cases, \([3k^2 + 5k + 2]\) would be even, so it can be represented as \(2h\), and therefore \([p^2 - 1 = (2h)*12]\). This means \([p^2-1]\) would again be a multiple of 24.