Find all positive integers n such that n, n + 2, and n + 4 are 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!.
+1262 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)
+26393 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 \(n=3\), \(n+2=5\), \(n+4= 7\)
