primes

If n = 3, then n + 2 = 5, and n + 4 = 7; all primes.

As far as I know there is only one such example. And that is:

3, 5, 7.  Other than this exception, every 3rd number in the form: n, (n + 2), (n + 4) is always divisible by 3. Here are a few examples of "twin primes":  (11, 13) , (17, 19) , (29, 31) , (41, 43) , (59, 61) , (71, 73) , (101, 103) , (107, 109) . In each of these examples, adding 2 to the 2nd prime or adding 4 to the first prime, it becomes divisible by 3. Always!.

so the prime numbers list under 100 is

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 and we have to have prime numbers that are 0,2, and 4 more than itself are only (3,5,7) so the answer (3,5,7)

Find all positive integers n such that n, n + 2, and n + 4 are all primes.

All prime numbers past 3 are of one of the form 6n+1 and 6n-1.

All integers are of one of this forms:
$\begin{cases}6n-2 & \Rightarrow& 2·(3n-1) \\ \mathbf{6n -1} \\6n & \Rightarrow & 2·3·n \\ \mathbf{6n+1} \\ 6n+2 & \Rightarrow& 2·(3n+1) \\ 6n+3 & \Rightarrow & 3·(2n+1)\end{cases}$

Note that all other than 6n-1 and 6n+1 can be expressed as a product of two integers bigger than 1.

So a prime number cannot be of any form other than $\mathbf{6n\pm 1}$

(That doesn't mean that all numbers of the form $\mathbf{6n\pm 1}$ are prime)

So $n=6n-1$, and $n+2 = 6n+2$ , we see $n+4 = 6n+3$ is always divisible by 3, so there is no more solution, if $n > 3$.

The only solution is $n=3$, $n+2=5$, $n+4= 7$