How do you prove that $n^2-1$ is composite when $n>2$? [duplicate]

When $n=2$ we know that $n^2-1$ is prime but is this the case, when $n>2$?

• Difference of two squares? – saulspatz Jul 1 '18 at 15:43
• $n^2 - 1 = (n-1)(n+1)$ – AlohaSine Jul 1 '18 at 15:43
• And also a duplicate of this. Closed as well. Hmm... – Jyrki Lahtonen Jul 1 '18 at 15:51
• But, this has survived! – Jyrki Lahtonen Jul 1 '18 at 15:52
• A veteran tip. Use Approach0! – Jyrki Lahtonen Jul 1 '18 at 15:53

HINT: $$n^2-1=(n-1)(n+1)$$ so it can not be prime