If 'n' is a prime number greater than 5,
then prove that (n^4 - 1) is divisible by 240?
My Solution
n = 2k+1, since all primes > 5 are odd.
n^4 - 1 =
(2k)^4 + 4(2k)^3 + 6(2k)^2 + 4(2k)
16k^4 + 32k^3 + 24k^2 + 8k = S
240 = 16 * 3 * 5
If k is even, S is divisble by 16 = 2 * 8.
If k is odd, 24 (k^2) + 8 k = 8 k (3k+1),
and 3k+1 is even, so S is divisible by 16 = 2 * 8.
Therefore S is divisible by 16.
k mod 3 = 0 => S divisible by 3
k mod 3 = 1 => (2k+1) = 0 mod 3 => n not prime
k mod 3 = 2 => (2k+1) = 2 mod 3, 2^4 = 1 mod 3,
so n^4 - 1 is divisible by 3.
n mod 5 = 0 => n not prime
n mod 5 = 1 => (n^4 - 1) = 0 mod 5
n mod 5 = 2 => (n^4 - 1) = 0 mod 5 (16 -1 = 15)
n mod 5 = 3 => (n^4 - 1) = 0 mod 5 (81 -1 = 80)
n mod 5 = 4 => (n^4 - 1) = 0 mod 5 (256 - 1 = 255)
Therefore n^4 - 1 divisible by 16, by 3, and by 5, so by 240.
Here is another proof by jgoulden
Let's factor ( n^4 - 1 ) as ( n^2 + 1 )( n - 1 )( n + 1 ).
Since n is odd, each term is divisible by 2.
Since n > 5, either n-1 or n+1 is divisible by 4.
Therefore, the entire expression is divisible by 16.
n is prime, so n % 3 is either 1 or 2.
If n % 3 = 1, then n - 1 is divisble by 3.
If n % 3 = 2, then n + 1 is divisible by 3.
n is prime, so n % 5 is 1, 2, 3, or 4.
If n % 5 = 1, then n - 1 is divisible by 5.
If n % 5 = 2 or 3, then n^2 + 1 is divisible by 5
If n % 5 = 4, then n + 1 is divisible by 5.
So our expression is divisible by 16, 3, and 5 and is thus
divisible by 240.
