Sum of two co-prime integers I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $\gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.

Here's another route you can take to solve this problem. For any $n \ge 7$, you want to show that there is a number $a$ where


*

*$gcd(a, n - a) = 1$,

*$1 < a < n$, and

*$1 < n - a < n$.


One option would be to choose $a$ to be the smallest prime number that doesn't divide $n$. In that case, $gcd(a, n - a) = 1$ because otherwise you'd have $gcd(a, n - a) = a$, meaning that $a$ divides $a + (n - a) = n$, contradicting the fact that $a$ doesn't divide $n$.
What you'll need to then show is that if you pick $n \ge 7$ that the smallest prime number that doesn't divide $n$ happens to be less than $n - 1$. I'll leave that as an exercise to the reader. :-)
