Hi
Sorry, I guess I will have to make a new topic.
Let n > 3 be a natural number. If n^2 + 2 is a prime number, then n is dividable
with number 3 - I want to prove this.
Here's my solution;
let's try with numbers which are dividable with 3.
6,9,12,15,18,21 and so on.
6^2 + 2 = 38 . but 38 isn't a prime number.
9^2 + 2 = 83 . this is.
So, 83 is a prime number and 9 is dividable with 3.
12 doesn't produce a prime number.
also 15^2 + 2 = 227, is a prime number and 15 is dividable with 3.
18 doesn't produce a prime number.
21 does produce and it is dividable with 3.
I don't know why some of these ( 6,9,12,15,18,21 and so on. )
numbers + 2 produce a prime number but not all?
Is this a good way to prove this?
Sorry, I guess I will have to make a new topic.
Let n > 3 be a natural number. If n^2 + 2 is a prime number, then n is dividable
with number 3 - I want to prove this.
Here's my solution;
let's try with numbers which are dividable with 3.
6,9,12,15,18,21 and so on.
6^2 + 2 = 38 . but 38 isn't a prime number.
9^2 + 2 = 83 . this is.
So, 83 is a prime number and 9 is dividable with 3.
12 doesn't produce a prime number.
also 15^2 + 2 = 227, is a prime number and 15 is dividable with 3.
18 doesn't produce a prime number.
21 does produce and it is dividable with 3.
I don't know why some of these ( 6,9,12,15,18,21 and so on. )
numbers + 2 produce a prime number but not all?
Is this a good way to prove this?