a) Find with explanation all triplets of primes of the form (n,n+2,n+4)
b) Find five evenly spaced primes - that is, five primes forming an arithmetic progression.
Notice that adding 2 to a number changes its remainder when divided by 3 by -1. That means the possible remainders (n, n+2, n+4) can leave when divided by 3 is: (-1, 1, 0), (0, -1, 1), and (1, 0, -1), so no matter what, at least one of (n, n+2, n+4) is divisible by 3. That means the only possibility is that n is 3, so the only triplet of primes in that form is (3, 5, 7)
Using a similar logic as above, we can tell that in a group of 5 evenly spaced numbers where the difference is not a multiple of 5, exactly one of them is divisible by 5.
That means you have to set the first term equal to five and add a non-multiple of 5 4 times until you get one of them that makes all the numbers prime. can you finish it?
(a) There will be a multiple of 3 if it starts with a multiple of 3.
There will be a multiple of 2 if it starts with a multiple of 2.
There will be a multiple of 5 if it starts with a multiple of 5.
There will be a multiple of 7 if it starts with a multiple of 7.
Thus, we only have to try (1, 3, 5), (2, 4, 6), (3, 5, 7), (5, 7, 9). The only triplet that works is (3, 5, 7).
(b) There are none. There will be a multiple of something if it gets bigger. We only try the first few, and see that they all include a number that is either 1 or composite.
5, 11, 17, 23, 29 works :)
also, quick correction, i said that the sequence has to start with a 5; it does not. It could start with a 7 (example : (7,13,19,25,31)) or 101 (example:101,9976511,19952921,29929331,39905741) ) or any prime number, but the difference has to be a multiple of 5 if the first term is not a 5.