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Find all numbers $p$ such that both $p, \ p^2+14$ are primes.

I believe only $ p= 3 $ works.

How do I prove this using the complete set of residues Modulo 3. Would something like this work.

Assume $\mathbb F = \{p, \ p+1, \ p+2 \} $ be the set of complete residues in $\mathbb Z/3 \mathbb Z$, then one of the elements of $\mathbb F$ is a zero. Meaning it is a multiple of $3$. This cannot be $p+1$ because then $p$ would equal $2$ and $4+14$ is a composite integer. Same for $p+2$ as $p = 1$ is not a prime. Hence $p = 3$.

I am still unsatisfied with this proof.

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1 Answer 1

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If the number $k$ is not a multiple of $3$, then $k=3n\pm 1$ for some $n$, and $$k^2=9n^2\pm6n+1=3(3n^2\pm2n)+1=3m+1$$ for some $m$, that is, $k^2\equiv1\pmod 3$ if $3$ does nor divide $k$, so $k^2+14\equiv 0\pmod 3$. In particular, if $3$ does not divide $p$, then $3$ divides $p^2+14>3$, so $p^2+14$ is not a prime.

On the other hand, if $p=3$, then $p^2+14=23$ is prime as well, so $p=3$ is the only solution.

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  • $\begingroup$ beautiful...thank you $\endgroup$
    – Quester
    Commented Nov 13, 2013 at 8:38

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