Prime numbers of the form $6k\pm 1$ I've got a conjecture:-
If $k$ is a natural number, then either $6k+1$ or $6k-1$ is a prime number.
Is this true?
 A: No it is not, say $120+1 = 11 ^2$ and $120-1 = 7\cdot 17$
A: According to elementary number theory, however, you can represent every prime number in the form $6k\pm1$ so your conjecture I would say is close to true however it does not mean that every number of the form $6k\pm1$ is a prime. This, however, holds true only for all primes $\gt3$. That is $2$ and $3$ are not written in this form.
A: The reason this looks as though it might be true is that $6k\pm 1$ excludes the possible small factors $2$ and $3$, so counterexamples will involve multiples only of $5, 7, 11, 13, 17, 19, 23, 29, \dots$.
Looking at admissible multiples of $5$ we have $25 (23), 35 (37), 55 (53), 85 (83), 95 (97), 115 (113), 145:143 (11\times 13)$ where numbers in brackets are prime.
John Watson has given an example which works for multiples of $7$.
Conjectures of this kind can look quite compelling if small primes are excluded - even just $2$ and $3$ in this instance, but careful analysis is required before they become truly credible.
