BMO1 2006 1

Note that for a prime, p>6:

p\equiv1,5 \pmod6

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:

n-1\equiv5\pmod6\\ n+1\equiv1\pmod6

so n is a multiple of 6 or n=6k for some integer k.

\begin{align} n^2(n^2+16)&=36k^2(36k^2+16)\\ &=144k^2(9k^2+4) \end{align}

so n^2(n^2+16) is divisible by 144.

Since n-1 and n+1 are not divisible by 5:

\begin{align} n&\equiv0,2,3\pmod5\\ n^2&\equiv0,4\\ \end{align}

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.

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These have already been solved but they are under 2005

Ah it sounds like I’ve made a mistake here, I shall remedy