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Prove that the only prime number $p$ for $p^2+2$ to be prime as well is $p=3$

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closed as off-topic by Pragabhava, suomynonA, Cameron Williams, Leucippus, TravisJ Oct 21 '16 at 2:39

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  • $\begingroup$ $p^2\equiv1\pmod 3$ $\endgroup$ – JonMark Perry Oct 21 '16 at 12:05
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If $p\neq 3$ then it has form $3k\pm 1$ so $p^2+2=(3k\pm 1)^2+2=9k^2\pm6k+3=3(3k^2\pm 2k+1)$ so wont be prime

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  • $\begingroup$ Why hasn't it got a form $3k \pm 2$ as well? $\endgroup$ – Bernard M. Oct 20 '16 at 15:45
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    $\begingroup$ @BernardM.: Notice that $3k+2 = 3(k+1) - 1$ and $3k-2 = 3(k-1) + 1$, so the formula $3k \pm 1$ covers all the cases different from $3k$. $\endgroup$ – Alex M. Oct 20 '16 at 15:53
  • $\begingroup$ Thanks a lot! Now I understand that $\endgroup$ – Bernard M. Oct 20 '16 at 16:08
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hint: If $q = p^2 + 2$ is a prime number, then $p = 3k+1$ or $p = 3k+2$

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    $\begingroup$ I believe that you meant to use $\ne$ instead of $=$. $\endgroup$ – Alex M. Oct 20 '16 at 15:47

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