Question

If $n$ is a prime number greater than $5$, show that $n^4-1$ is divisible by $240$.

Answer 1

$n^4-1$

Using fermats little theorem if $p$ is prime and $N$ is prime to $p$ then $N^{p-1}-1$ is divisible by $p$

Put $n=5$

$\Rightarrow n^{5-1}-1=n^4-1$ is divisible by $\mathrm{5.....}\left(i\right)$

$n^4-1=\left(n^2-1\right)(n^2+1)=(n-1)(n+1)(n^2+1)$

Again using fermats theorem

Put $n=3$

$\Rightarrow n^{3-1}-1=n^2-1$ is divisible by $\mathrm{3......}\left(ii\right)$

As $n$ is prime greater than $5$ so $n$ is odd

$\Rightarrow (n^2+1)$ is even which is divisible by $\mathrm{2.....}\left(iii\right)$

Now $(n-1)(n+1)$ is product of two even consecutive 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 $\mathrm{8.........}\left(iv\right)$

So by using $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)$

$n^4-1$ is divisible by $5\times 3\times 8\times 2=240$

Hence proved.