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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 $n^2 - a$?

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    $\begingroup$ Use that $$n^2-1=(n-1)(n+1)$$ and this can not be a prime for $n>2$ $\endgroup$ Commented Jul 3, 2019 at 19:54
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    $\begingroup$ $n^2-1=(n-1)(n+1)$ so can it be a prime in general??? $\endgroup$
    – Anurag A
    Commented Jul 3, 2019 at 19:54
  • $\begingroup$ Use the polynomial $$n^2-n+41$$ to generate more primes. $\endgroup$ Commented Jul 3, 2019 at 19:56
  • $\begingroup$ @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. $\endgroup$
    – Anurag A
    Commented Jul 3, 2019 at 19:58

2 Answers 2

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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. $n = 2$.

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Welcome to the Mathematics StackExchange!

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).

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