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For which primes $p$, $p+10$ and $p+14$ are also primes?

I assume it has something in common with division (whether the prime $n$ is divisible by some number), but that is just the first idea that came to my mind - could you help?

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2 Answers 2

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Lets assume $p\not=3$. Then there are 2 cases: $p=3k+1$ and $p=3k+2$ for some $k\in\mathbb{N}$. If $p=3k+1$, then $p+14=3k+15=3(k+5)$, which means it's divisible by $3$ which is a contradiction. Now if $p=3k+2$, then $p+10=3k+12=3(k+4)$ which is divisible by $3$. So only case is when $p=3$. Now those are $3,13,17$ which are prime.

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Note that one of $p+10, p+12, p+14$ must be divisible by $3$ (three consecutive members of an integer arithmetic progression where the common difference is not divisible by $3$), so one of $p+10, p, p+14$ is divisible by $3$.

The only prime multiple of $3$ is $3\lt 10$, so the only possibility is $13, 3, 17$

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