When $n=2$ we know that $n^2-1$ is prime but is this the case, when $n>2$?
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Difference of two squares? – saulspatz Jul 01 '18 at 15:43
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2$n^2 - 1 = (n-1)(n+1)$ – MathematicsStudent1122 Jul 01 '18 at 15:43
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And also a duplicate of this. Closed as well. Hmm... – Jyrki Lahtonen Jul 01 '18 at 15:51
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But, this has survived! – Jyrki Lahtonen Jul 01 '18 at 15:52
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4A veteran tip. Use Approach0! – Jyrki Lahtonen Jul 01 '18 at 15:53
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Voting on the question was perhaps a bit harsh (closing was inevitable). Not much I can do about that, sorry. – Jyrki Lahtonen Jul 01 '18 at 16:21
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Sorry that I wasted your time! – Kfkcigic Jul 01 '18 at 17:10
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Business as usual :-) Don't worry about it. Better luck with the next question. Do check out the local guide! And try to search! – Jyrki Lahtonen Jul 01 '18 at 21:15
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Thank you. I will – Kfkcigic Jul 02 '18 at 06:56
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HINT: $$n^2-1=(n-1)(n+1)$$ so it can not be prime
Dr. Sonnhard Graubner
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Thank you! I was so stupid! I should have noticed that factoring formula. Sorry that I wasted your time. – Kfkcigic Jul 01 '18 at 17:04