I recently stumbled upon a fact that all prime numbers past $3$ are of the form either $6n-1$ or $6n+1$. Is it true? at least for numbers less than $10^9$. And does it cover all primes?

Jan 24, 2015 · Prove there are infinitely many primes of the form $6n - 1$ with the following: (i) Prove that the product of two numbers of the form $6n + 1$ is also of that form.

My own solution: According to the division algorithm a number can be of the form $6n$, $6n+1$, $6n+2$, $6n+3$, $6n+4$ or $6n+5$.

Mar 1, 2018 · A number of the form $6n+5$ is not divisible by $2$ or $3$. Now note that the product $ (6n+1) (6m+1)=36nm+6n+6m+1=6 (6mn+m+n)+1$, and you can show by induction …

There should be infinitely many primes of the form $5+6n$. How do you prove it? The same should be true for $7+6n$.

6n = 149 - 1 n = 148 / 6 n = 24.666~7 6n - 1 = 149 6n = 149 + 1 n = 150 / 6 n = 25 If you notice both pass Euler's test under the evaluation of 6n-1. Am I oversimplifying Euler's theorem as …

Feb 4, 2020 · Here it is: "It seems that the sequence of all integers N that are either 6n+1 or 6n−1 are all primes if N is neither divisible by 5 nor the product of primes greater than or equal to 7. …

May 28, 2020 · I was stuck on a problem from Mathematical Circles: Russian Experience, which reads as follows: Prove that the number $6n^3 + 3$ cannot be a perfect sixth power of an …

Nov 29, 2015 · Let m be a integer. Then if $6n+1$ is a composite number we have that $\\operatorname{lcd}(6n+1,m)$ is not just $1$, because then $6n+1$ would be prime. Also …

Sep 30, 2020 · Looking at the Mathworld entries on these theorems here and here, I notice that representation of primes of the form $4n+1$ is stated to be unique (up to order), but that there …