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I need help with this question: There are infinitely many prime numbers $p$ for which $p + 2$ and $p + 4$ are also primes.

Where should I start? what proof techniques will be useful?

Any input will be very useful to me.

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    At least one of $p, p+2, p+4$ are multiple of 3, so your conjecture (or the question) is not true. – 2017-02-10
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    how do disprove it then? – 2017-02-12
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    Possible duplicate of [How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$](https://math.stackexchange.com/questions/522585/how-can-i-prove-that-one-of-n-n2-and-n4-must-be-divisible-by-three-fo) – 2017-10-14
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    @kingW3 Yeah, I should have known this was a duplicate, too. – 2017-10-16

2 Answers 2

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If $p>3$ is a prime then reducing mod 3, one of $p+2$ and $p+4$ is $0\pmod 3$ and so is divisible by $3$. Thus, the only prime $p$ for which both $p+2$ and $p+4$ are also prime is $p=3$.

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    I am not supposed to use mod and congruence in this section to answer this question. I need to use the division algorithm somehow tho. Any ideas? – 2017-02-12
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No, there are not infinitely many such primes among the integers, $\mathbb Z$. In fact, there are only precisely two of them: $p = -7$ or 3.

With the former, we have $p + 2 = -5$ and $p + 4 = -3$. With the latter, we have $p + 2 = 5$ and $p + 4 = 7$. So we have either the three primes $-7, -5, -3$ or the three primes 3, 5, 7. This suggests that in such a triple, one of the numbers must be a multiple of 3.

Since you can't use congruences, the solution is somewhat more laborious, but not by much. Suppose $p = 3k + 1$. Then $p + 2 = 3k + 3 = 3(k + 1)$, and that's composite unless $k = -2$ or 0. So suppose instead that $p = 3k + 2$. Then maybe $p + 2$ is prime, but then $p + 4 = 3k + 6 = 3(k + 2)$. The case $p = 3k$ (with implied plus zero) should now be obvious.