We know that any prime number ( $> 2,3$) can be written in the form $6k+1$ or $6k-1$. Is it necessary that at least one of $6k+1$ or $6k-1$ is a prime number ?
2026-04-09 06:02:45.1775714565
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Is at least one of $6k + 1$ or $6k-1$ prime?
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No, for lots of reasons:
There are arbitrarily long sequences of composite numbers. In fact, to construct a sequence of length at least $N - 1$, consider $N! + 2, ..., N! + N$. Letting $N$ be at least $9$ guarantees that this is very false.
A counterexample can be found by counting for a while.
If this were true, it would guarantee that there are infinitely many prime gaps of size at most $8$. This would be a significant improvement on recent work by Yitang Zhang.
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Any number must in in this series $6k-3,6k-2,6k-1,6k,6k+1,6k+2$
$k=1$ then $3,4,5,6,7,8$
$k=2$ then $9,10,11,12,13,14$
$k=3$ then $15,16,17,18,19,20$
$k=4$ then $21,22,23,24,25,26$
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In that series, $6k-3,6k-2,6k,6k+2$
cannot be prime because they are composite numbers. $3(2k-1),2(3k-1),6k,2(3k+1))$
You have only 2 options for prime numbers. $6k-1$ or $6k+1$ but we do not know if they are prime numbers or not. We can only think that they are candidate for prime number. As barak manos pointed,
$k=20$ then $117,118,119,120,121,122$
Our prime candidates in the form of $6k-1,6k+1$ are $119,121$ but both of them are composite numbers.
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Constructing $k$ that are counter-examples is easy, by means of choosing what number should be a divisor of $6k-1$ and what number should be a divisor of $6k+1$.
If we decide to have $m | (6k-1)$ and $n \mid (6k+1)$, then that is the same thing as saying
$$ 6k \equiv 1 \pmod{m} $$ $$ 6k \equiv -1 \pmod{n} $$
As long as $m,n,6$ are all relatively prime, we can solve this system of equations for $k$; e.g. by using the Chinese Remainder Theorem.
No, take $k=20$ for example:
As a side note, if the answer to your question was yes, then it would imply that the total amount of prime numbers is somewhere between $\displaystyle\frac{1}{6}$ of all positive integers to $\displaystyle\frac{1}{3}$ of all positive integers.