The other week, a colleague walked into my office and posed the following problem to me:
For all n where n is prime and n > 3, show that n² – 1 is divisible by 24.
Examples: 5² – 1 = 24; 7² -1 = 48; 11² – 1 = 120.
(If you’re interested, give it a go before looking at the solution. It doesn’t require any advanced Maths.)
First, what does it mean to be divisible by 24? Well, it means being divisible by 2, 2, 2, and 3 (the prime factorization of 24). So we have to find a factor of 3, and 3 factors of 2.
As any highschool mathematician knows, n² – 1 = (n + 1)(n – 1).
Since n is prime and n > 3, it must be odd. Which means that both (n + 1) and (n – 1) must be even (divisible by 2). Moreover, since they are consecutive multiples of 2, one of them must be divisible by 4, and the other by 2.
We also know that (n – 1), n, (n + 1) are three consecutive numbers, so one of them must be divisible by 3. Since n is prime and n > 3, it cannot be divisible by 3. So either (n – 1) or (n + 1) must be divisible by 3.
Which means that (n + 1)(n – 1) is divisible by 24. QED.