Let n be an integer greater than 6. Prove that if n-1 and n+1 are both prime, then n2(n2+16) is divisible by 720.
(In reply to
How about ... by Larry)
Well, that is not completely right. Lets fix up Larry's proof.
If n-1 and n+1 are both prime, then n is even.
Since neither n-1 nor n+1 is divisible by 3, n is divisible by 3.
Therefore 6 divides n.
So 36 divides n^2.
(n^2+16) is divisible by 4; so we're up to 144
We need one more factor of 5.
N cannot end in 4, because then N+1 is a multiple of 5 (so not prime)
N cannot end in 6, because then N-1 is a multiple of 5 (so not prime)
If N ends in 0, then N^2 ends in 0.
If N ends in 2 or 8, N^2 ends in 4 and adding 16 makes a final digit of 0.
So the expression is divisible by 5 for any of the three possible final digits of N.
q.e.d.
re: How about ...
