Are there infinitely many primes of the form $n^2 - 1$?

By computation, I feel like there is a finite number of prime (the only prime I found is where $$n = 2$$, so $$n^2 -1 = 3$$)

Also, for the general form $$n^2 - a$$ where a is some positive integer:

For which values of a will there be an infinite number of primes given by $$n^2 - a$$?

• Use that $$n^2-1=(n-1)(n+1)$$ and this can not be a prime for $n>2$ Commented Jul 3, 2019 at 19:54
• $n^2-1=(n-1)(n+1)$ so can it be a prime in general??? Commented Jul 3, 2019 at 19:54
• Use the polynomial $$n^2-n+41$$ to generate more primes. Commented Jul 3, 2019 at 19:56
• @Dr.SonnhardGraubner I don't think your second comment is related to OP's question because OP is looking for $n^2-a$, where (most likely) $a$ is some fixed integer. Commented Jul 3, 2019 at 19:58

We have $$n^2 - 1 = (n - 1)(n+1)$$. Therefore the number $$n^2 -1$$ is only prime, when one of these factors equals $$1$$, i.e. $$n = 2$$.
Given that $$n^2-a=(n+\sqrt a)(n-\sqrt a)$$, we can already rule out that there could exist infinitely many primes of the form $$n^2-a$$ if $$a$$ is a square of a natural number. Other than that, there seems to be no obvious bound to the number of primes of this form (even if, say, $$n^2-2$$ will never be prime for even $$n$$, this still leaves infinitely many candidates).