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Is there any integer $p$ such that $p > 1$ and all three numbers $p$, $p+2$ and $p+4$ are prime numbers? If there are such triples, prove that you have all of them; if there are no such triples, prove why not.

Apart from this, I am given the following information: if $a$ is an integer, then precisely one of the following holds: either $a = 3k$ for some integer $k$, or $a = 3k+1$ for some integer $k$, or $a = 3k+2$ for some integer $k$.

I have observed that, when you sum $p$, $p+2$ and $p+4$, you get $3(p+2)$. But from that, I don't know how to proceed.

Y.T.
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user2812
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3 Answers3

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If $p$ is not divisible by $3$, then either $p+1$ or $p+2$ must be. (Otherwise, there would be 3 consecutive numbers that are not divisible by $3$, which makes no sense)

If $p+2$ is divisible by $3$, then clearly it is not prime, unless it is $3$. But that would mean that $p=1$, which isn't prime.

But if $p+1$ is divisible by $3$, then $p+4$ must also be, as $p+4=(p+1)+3$. So then $p+4$ is not prime, unless it is $3$. But then $p$ would be $-1$, which is not prime.

So, $p$ must be divisible by $3$, because otherwise, one of the two other numbers is. But if $p$ divides $3$, it must be composite, unless it is $3$.

Therefore, the only three numbers that work are $3,5,7$.

Polygon
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Note that $p(p+2)(p+4)=p(p+1)(p+2)+3p(p+2)=3!{p+2 \choose 3}+3p(p+2)$ is a multiple of $3$. Therefore, at least one of $p,p+2,p+4$ must be a multiple of $3$, and hence equal to $3$, since each is a prime. The only option is for $p=3$, since $p>1$. We check that $3,5,7$ are all primes. $\blacksquare$

AT1089
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Suppose that $p$ is prime and that $p>3$, then either $p\equiv 1 \space(\text {mod 3})$ or $p\equiv 2 \space(\text {mod 3})$.

If the first case, then $p+2$ is a multiple of $3$ and if the second case, then $p+4$ is a multiple of $3$. Then it is clear that for neither $p+2$ nor $p+4$ to be composite that $p$ must be a multiple of $3$. Since, $p$ is prime the only solution is $p=3$.

This goes beyond the scope of your question but in general if you have a tuple of positive integers $(p,p+2^n,p+2^{n+1})$, where $n>0$ and you want all $3$ numbers to be prime then the only potential solution is $p=3$. This is because $4^m\equiv 1 \space(\text {mod 3})$ and $2^{2m+1}\equiv 2 \space(\text {mod 3})$, where $m\geq0$.

Robotichate
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