Question

Show that the difference of the squares of any two prime numbers greater than $6$ is divisible by $24$.

Answer 1

Let $m$ and $n$ be two prime numbers greater than $6$

Consider another prime number $N$

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 $6$ 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 $8$

Now given number is $=m^2-n^2$

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

Both parts of given number are of form $N^2-1$

So they will be divisible by $24$

Hencce proved.