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

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  • $\begingroup$ Difference of two squares? $\endgroup$ – saulspatz Jul 1 '18 at 15:43
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    $\begingroup$ $n^2 - 1 = (n-1)(n+1)$ $\endgroup$ – AlohaSine Jul 1 '18 at 15:43
  • $\begingroup$ And also a duplicate of this. Closed as well. Hmm... $\endgroup$ – Jyrki Lahtonen Jul 1 '18 at 15:51
  • $\begingroup$ But, this has survived! $\endgroup$ – Jyrki Lahtonen Jul 1 '18 at 15:52
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    $\begingroup$ A veteran tip. Use Approach0! $\endgroup$ – Jyrki Lahtonen Jul 1 '18 at 15:53
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HINT: $$n^2-1=(n-1)(n+1)$$ so it can not be prime

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  • $\begingroup$ Thank you! I was so stupid! I should have noticed that factoring formula. Sorry that I wasted your time. $\endgroup$ – Kfkcigic Jul 1 '18 at 17:04

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