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  1. Show that if $ n ^ 2 + 2 $ is prime, then $ 3 \mid n $.
  2. Show that the only prime of the form $n ^ 3-1$ is $7$.

For the 2. Let $ p = n ^ 3-1 $. Let's see that the only possibility for $ p = 7.$ For that note that $ n ^ 3-1 $ can be decomposed as follows. $$ n ^ 3-1 = (n-1) (n ^ 2 + n + 1). $$ Since $ p $ is prime, and $ p = n ^ 3-1 = (n-1) (n ^ 2 + n + 1) $, then there are two possibilities.

  1. Possibility 1. That $ p = n-1 $ and $ n ^ 2 + n + 1 = 1 $. Hence it will follow that $ n = -1 $ or $ n = 0 $. But if we replace $ n = -1 $ or $ n = 0 $, then $ p $ would not be prime.

  2. Possibility 2. That $ p = n ^ 2 + n + 1 $ and $ n-1 = 1 $. From this it follows that $ n = 2 $. Therefore, $$ p = 2 ^ 2 + 2 + 1 = 7. $$

Therefore we can conclude that the only prime that of the form $ n ^ 3-1 $ is $ 7 $, which ends the proof.

but for the $ 1 $ I have no idea how to do it, any help?

Thomas Andrews
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asd asd
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    Since the questions are not directly related, you should ask them separately. They're free! :) (Also: Please give questions informative titles.) – Blue Mar 27 '21 at 15:25
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    @Blue In fact the OP already gives a solution to the second question. I guess it is presented as context, showing what the OP's level is. Thus it's OK: there is just one question. – WhatsUp Mar 27 '21 at 15:31
  • @asd asd - I assume you mean positive prime? I say this because $n=-1$ yields $p=-2$ in your possibility $1.$ – Chris Leary Mar 27 '21 at 15:36
  • Yes, we only work with positive prime numbers. – asd asd Mar 27 '21 at 15:37
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    $n=1$ gives $n^2+2=3$ which is a prime but $3 $ does not divide $1$. I've added an answer to include this case. – Aditya Mar 27 '21 at 15:44

2 Answers2

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If $n$ is not a multiple of $3$ then it can be

$n=3k+1$ or $n=3k+2$

in both cases $n^2+2$ is not prime. Contradiction.

Therefore $n$ is a multiple of $3$.

Raffaele
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Actually for (1), $n=1$ is a counterexample (trivial case, but counterexample nonetheless).

$n^2 \equiv 0$ or $1 \pmod 3$.

$\implies n^2+2\equiv 2$ or $0 \pmod 3$ and $n^2+2 \neq 0 \pmod 3$ (and here is the trivial case, where $n^2+2=3 \implies n=1$ in the positive integers)

Other than that one case $n^2+2$ must be $2 \pmod 3$, so $n^2 \equiv 0 \pmod 3 \implies n \equiv 0 \pmod 3 \implies 3|n.$

Aditya
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