Question

If $n-1,n+2$ are both prime numbers greater than $5$, shew that $n(n^2-4)$ is divisible by $120$, and $n^2(n^2+16)$ by $720$. Also show that $n$ must be of the form $30t$ or $30t\pm 12$.

Answer 1

Since $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!$ or $120$; and $(n-1)$ and $(n+1)$ are both primes and greater than $5$,

$\therefore n(n-2)(n+2)$is divisible by $120$, that is $n(n^2-1)$ is divisible by $120$. Again $(n-1)n(n+1)$ is divisible by $6$, and therefore n is divisible by $6$ since $(n-1)$ and $(n+1)$ are prime and greater than $5$

$\therefore n^2(n^2-4)$ is divisible by $720$, also $20n^2$ is divisible by $720$;

$\therefore n^2(n^2-4)+20n^2;$ or $n^2(n^2+16)$ is divisible by $720$

Lastly, $n=6s$ and one of the three members $n+2,n,n-2,$ is divisible by $5$

$1)n=6s=5r-2;s+\frac{s+2}{5}=r;s=5t-2$

$\therefore n=30t-12$

$2)n=6s=5r;n=30t$

$3)n=6s=5r+2;s=\frac{s-2}{5}=r;s=5t+2$

$\therefore n=30t+12$

$n=30t$ or $30t\pm 12$