Question

If $n$ is a prime number greater than $3$, show that $n^2-1$ is a multiple of $24$.

Answer 1

$n^2-1$

Using fermats little theorem if $p$ is a prime number and $N$ is prime to $p$ then $N^{p-1}-1$ is a multiple of $p$

$n^2-1=n^{3-1}-1$

As $n$ is prime and $3$ is also prime

$\therefore$ $n^2-1$ is a mutiple of $\mathrm{3.......}\left(i\right)$

$n^2-1=(n-1)(n+1)$

As $n$ is prime greater than $3$ so its is an odd integer.

$\Rightarrow$ $(n-1)$ and $(n+1)$ are two consecutive even integers.

So they will be of from $2n$ and $2n+2$

$\Rightarrow 2n(2n+2)\Rightarrow 4n(n+1)$

as $n$ is odd so $(n+1)$ is a multiple of $2$

$\Rightarrow 4n(n+1)$ is a mutiple of $8$

$\Rightarrow (n-1)(n+1)$ is a mutiple of $\mathrm{8.........}\left(ii\right)$

Hence using $\left(i\right)$ and $\left(ii\right)$ it is clear that $n^2-1$ is a mutiple of $8$