How to prove that the number 6+2 can be represented as the sum of two coprime numbers?

Question

$k$ is a natural number. How can I prove that the number $6k+2$ can be represented as a sum of two coprime numbers?

The greatest common divisor of two coprime numbers is $1$: For integers $m,n$, $\gcd(a,b) = ma+nb = 1$.

But what can I do from here? (The answer can't be: $a = 6k+1$, $b = 1$.)

The tools for the answer can be the division theorem, Bézout's identity, or Bézout's Lemma.

Answer 1

You can write $6k + 2 = (2k + 1) + (4k + 1)$. The two numbers on the right-hand side are relatively prime for any $k \in \mathbb{N}$ since $2(2k + 1) - (4k + 1) = 1$.