Note that for a prime, p>6:
as p is not a multiple of 2 or 3.
Since the difference between n-1 and n+1 is 2, it must be the case that:
so n is a multiple of 6 or n=6k for some integer k.
so n^2(n^2+16) is divisible by 144.
Since n-1 and n+1 are not divisible by 5:
Therefore, either n^2 or n^2+16 is divisible by 5, so n^2(n^2+16) is divisible by 5.
This means that n^2(n^2+16) is divisible by \gcd(144,5)=720.
For the converse, consider n=48. Since n-1 and n+1 are not divisible by any of 2,3 or 5, the above reasoning holds and n^2(n^2+16) is divisible by 720. However, n+1=49=7^2, so the converse is not true.
These have already been solved but they are under 2005
Ah it sounds like I’ve made a mistake here, I shall remedy