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Let $p$ be a prime number for which also $p^2+2$ is a prime. Show that then $p^3+2$ is also a prime.

Computing few first primes I got:

$p=2$: $2^2+2=6$ (not satisfying the condition)

$p=3$: $3^2+2=11$ (satisfying the condition)

$p=5$: $5^2+2=27$ (not satisfying the condition)

$p=7$: $7^2+2=51$ (not satisfying the condition)

so I would have a reason to believe that only the case $p=3$ will satisfy this condition, but how would I go about showing this rigorously?

Batominovski
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3 Answers3

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The only number satisfying this condition is $3$, for all other prime numbers we have: $$p \equiv \pm 1\pmod 3 \Rightarrow p^2 + 2 \equiv 0\pmod 3,$$ and therefore $p^2 + 2$ is not a prime number.

Kenta S
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Dmitry
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Except $3$ all the primes are of the form $3k+1$ or $3k+2$.

If $p$ is of the form $3k+1$ then, $p^2+2$ is divisible by $3$. If $p$ is of the form $3k+2$, then $p^2+2$ is again divisible by $3$. Hence there are no primes satisfying both conditions, except $3$.

mertunsal
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If $3 \nmid p$, then $p \equiv \pm 1\pmod{3}$ and so $3 \mid (p^2+2)$, with $p^2+2>3$. So $p^2+2$ is prime can only hold for $p=3$. We note that both $3^2+2$ and $3^3+2$ are primes. $\blacksquare$

AT1089
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