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I came across a question on brilliant.org where it had been mentioned that $X, X + 2, X + 4$ are all prime numbers for $X = 3$. So is there any other $X$ for which $X, X + 2, X + 4$ are all primes and the answer was no.

So, can we say that there cannot be infinite number of prime triplets separated by $2$.

Please explain why?

David R.
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Aashish Loknath Panigrahi
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2 Answers2

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At least one of them is divisible by $3$. (Indeed, I suppose exactly one of them is divisible by $3$.)

Patrick Stevens
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As mentioned by others, one of them will be divisible by 3.

To demonstrate this, we assume that $x$ is not divisible by 3, so it will either be of the form $x = 3q+1$ or $x = 3q+2$ for some $q \in \mathbb N$.

If $x = 3q+1$, then $x+2 = 3q+3$, so $x+2$ is divisible by 3. If, on the other hand, $x = 3q+2$, then $x+4 = 3q+6$, and thus divisible by 3.

Therefore exactly one of the three numbers $x$, $x+2$ or $x+4$ will be divisible by 3, so they can only all be prime if $x = 3$.

BDN
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