One Less Than a Prime Squared

Notice that $p^2-1=(p-1)(p+1)$p2−1=(p−1)(p+1). Since $p$p is bigger than 3, $p$p must be odd, so $p-1$p−1 and $p+1$p+1 must be even numbers. Since $p-1$p−1 and $p+1$p+1 are consecutive even numbers, one of them is divisible by 4. From this, we can conclude that $p^2-1$p2−1 is divisible by $2\cdot 4=8$2⋅4=8.

Let’s consider the numbers $p-1$p−1, $p$p, and $p+1$p+1. Since they are 3 consecutive numbers, one of them must be a multiple of 3. Since $p$p is a prime that is greater than 3, the multiple of 3 must be either $p-1$p−1 or $p+1$p+1. This means that $p^2-1$p2−1 is also divisible by 3.

Since 8 and 3 share no factors in common, $p^2-1$p2−1 must be divisible by $8\cdot 3=24$8⋅3=24.