Prove that the only prime of the form $n^2− 1$ is 3, where $n$ is a natural number.
We have
$$n^2-1 = (n+1)(n-1) $$
The only way $(n+1)(n-1)$ is prime and not a composite number is if either $(n+1)$ or $(n-1)$ is 1
Let's consider each case.
- Case $(n+1)=1$ means $n=0$ in which case $(n+1)(n-1)$ is 0, and not a prime.
- Case $(n-1)=1$ means $n=2$ in which case $(n+1)(n-1)=3$.
So the only prime of the form $n^2-1$ is 3.