I'm trying to figure out how to prove that if $p$ and $p^2+2$ are prime numbers then $p^3+2$ is a prime number too. Can someone help me please?
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2The trick is that $p$ and $p^2 + 2$ are almost never both prime. Have you tried some examples? – Qiaochu Yuan Oct 06 '15 at 07:01
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Hint: Find what is $p\mod3.$ – CIJ Oct 06 '15 at 07:02
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@QiaochuYuan what about 3, 3 is prime and 11 is prime and 29 is prime ! – Nizar Oct 06 '15 at 07:04
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Prime numbers greater than $3$ are always $\pm1\bmod 6$. So $p^2+2\equiv 3\bmod 6$, and hence a multiple of $3$ (meaning not prime). – gebruiker Oct 06 '15 at 07:08
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1@Nizar: I said "almost never," not "never." – Qiaochu Yuan Oct 06 '15 at 07:10
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The two statements "almost never" and "never" are equivalent mod 3 – Euler....IS_ALIVE Oct 06 '15 at 07:13
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Related : http://math.stackexchange.com/questions/269790/why-does-p28-prime-imply-p34-prime – lab bhattacharjee Oct 06 '15 at 08:08
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If $p=2$, then $p^2+2$ is not prime.
If $p=3$, then $p^2+2 = 11$, then $p^3+2=29$ is prime.
If $p>3$, then $p \equiv \pm 1 \pmod 3$, then $p^2+2 \equiv 0 \pmod 3$. So, $p^2+2$ is not prime.
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For clarity you might want to add that the statement given by the OP is thus true... – gebruiker Oct 06 '15 at 07:13
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2Why do you bother to isolate the case $p = 2$ -- the logic holds for the same reason in the $p > 3$ case. – MT_ Oct 06 '15 at 07:47
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You can use the difference of squares formula to get that $p^2 + 2 = (p+1)(p-1) + 3$. Since $p^2$ is prime, $(p+1)(p-1)$ cannot have a factor $3$ as it would imply that $p^2 + 2$ is divisible by $3$ hence not a square. Among three consecutive integers one has to be divisible by $3$. Since neither $p+1$ or $p-1$ is divisible by $3$, $p$ must be. $p$ is a prime number and the only prime divisble by $3$ is $3$ itself, hence $p =3$ and $p^3 + 2 = 29$, a prime.
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