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The only prime $p$ such that $p+1$ is a perfect square is $3$

Which theorem I nee to use?

user568963
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    Just remember that $,a^2-b^2=(a-b)(a+b),$. – dxiv Jun 14 '18 at 04:21
  • There isn't really any theorem. Just note $p + 1 =k^2$ and hopefully you'll realize that that means $p = k^2 -1$ and it'll put you in mind of attempting to factor $k^2 - 1$ and seeing what conditions are required if $p$ is prime. – fleablood Jun 14 '18 at 05:11

1 Answers1

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If $p+1$ is a perfect square, we have $p+1 = n^2$ i.e., $p = n^2-1 = (n+1)(n-1)$.

If $n>2$ then $n-1$ $> 1$ i.e., p is not a prime.

Thus, the only case where $p+1$ can be a perfect square is if $n = 2$ i.e., $p = 3.$

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