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I need to prove that there is only one $p$ prime number such that $p^2+8$ is prime and find that prime.

Anyway, I just guessed and the answer is 3 but how do I prove that?

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Any number can be written as $6c,6c\pm1,6c\pm2=2(3c\pm1),6c+3=3(2c+1)$

Clearly, $6c,6c\pm2,6c+3$ can not be prime for $c\ge 1$

Any prime $>3$ can be written as $6a\pm 1$ where $a\ge 1$

So, $p^2+8=(6a\pm 1)^2+8=3(12a^2\pm4a+3)$.

Then , $p^2+8>3$ is divisible by 3,hence is not prime.

So, the only prime is $3$.


Any number$(p)$ not divisible by $3,$ can be written as $3b\pm1$

Now, $(3b\pm1)^2+(3c-1)=3(3b^2\pm2b+c)$.

Then , $p^2+(3c-1)$ is divisible by 3

and $p^2+(3c-1)>3$ if $p>3$ and $c\ge1$,hence not prime.

The necessary condition for $p^2+(3c-1)$ to be prime is $3\mid p$

$\implies$ if $p^2+(3c-1)$ is prime, $3\mid p$.

If $p$ needs to be prime, $p=3$, here $c=3$

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  • $\begingroup$ thank you but why any prime bigger than 3 can be write this way? $\endgroup$ – Mary Nov 10 '12 at 9:17
  • $\begingroup$ @Nusha: Because any other number is divisible by 2 or 3. Check out the cases $6n$, $6n+2$, $6n+3$, $6n+4$. (And note that $6n+5$ has the form $6n'-1$ with $n'=n+1$.) $\endgroup$ – Harald Hanche-Olsen Nov 10 '12 at 9:19
  • $\begingroup$ @Nusha, may have a look into the edited answer. $\endgroup$ – lab bhattacharjee Nov 10 '12 at 9:21
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    $\begingroup$ It's a fine answer, but if I may be allowed to editorialize a bit on grammar? The phrase “both factors $3$ and $…+3>1$ grates on my nerves. The phrase lists two items, $3$ and $…+3>1$ as if they were of a similar nature, which they are not. Please, either “both factors $3$ and $…+3$ are greater than $1$”, or “both $3>1$ and $…+3>1$”. $\endgroup$ – Harald Hanche-Olsen Nov 10 '12 at 9:24
  • $\begingroup$ @HaraldHanche-Olsen, you can edit it if you want $\endgroup$ – lab bhattacharjee Nov 10 '12 at 9:27
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Suppose there exists a prime $p$ (not equal to 3) such that $p^2 + 8$ is prime. Since $p$ is indivisible by 3, therefore $p \equiv 1 \pmod 3$ or $p \equiv -1 \pmod 3$, therefore $p^2 \equiv 1 \pmod 3$. Thus, $p^2 + 8 \equiv 9 \equiv 0 \pmod 3$, therefore $p^2 + 8$ is a prime greater than 3 that is divisible by 3 (a contradiction).

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Put $\rm\,q,k = 3\:$ in: $ $ for $\rm\,p\ne q\:$ primes, little Fermat $\rm\,\Rightarrow\, q\:|\: p^{q-1}\!-1+qn\ $ so it is prime iff it $\rm = q.$

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$p^2+8=0\pmod 3$ whenever $p=1,2\pmod 3$. For primes $p>3$, it is obvious that $p=1,2 \pmod 3$, and $p^2+8>3$, therefore $3$ is the only possible solution.

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