Number of $(a,b) \in \mathbb{N}^2$ such that : $\gcd(a,b) = 1$ and $a+b =n$ I would like to know : How many couples$ (a, b) \in \mathbb{N}^2$ are such that : 
$$a+b = n$$
with $n$ a fix positive integer and $a, b$ such that : $\gcd(a,b) = 1$
I really don't know how to proceed yet here are some values :
For example if with note : $Q_n$ the number of such couples for a certain $n$ we have :


*

*$Q_2 = 1$ because $1+1 = 2$

*$Q_3 = 2$ because : $3 = 1+2 = 2+1$

*$Q_4 = 2$ because $4 = 3+1 = 1+3$
 A: Note that whenever $(a,b)\in\mathbb{N}^2$ satisfy $a+b=n$, we have $\gcd(a,n)=\gcd(a,a+b)=\gcd(a,b)$, and thus.
$$\gcd(a,b)=1\iff\gcd(a,n)=1.$$
We must have that $1\leq a\leq n-1$ if $(a,b)\in\mathbb{N}^2$ satisfy $a+b=n$. Moreoever, for fixed $n\in\mathbb{N}$, if we are given an $a\in\mathbb{N}$ with $1\leq a\leq n-1$, we can find a unique $b\in\mathbb{N}$ with $a+b=n$ (simply take $b=n-a$.
Therefore, the number of pairs $(a,b)\in\mathbb{N}^2$ satisfying $a+b=n$ with $\gcd(a,b)=1$ is the same as the size of the set $\{a\in\mathbb{N} : 1\leq a < n,\ \gcd(a,n)=1\}$.
The Euler totient function is defined by
$$\phi(n) = |\{a\in\mathbb{N} : 1\leq a\leq n, \ \gcd(a,n)=1\}|$$
which agrees with the previous quantity, except when $n=1$ (because this is the only time where $a=n$ satisfies $\gcd(a,n)=1$). Hence the quantity you seek is $\phi(n)$, except when $n=1$, then the answer is 0.
A: As @lulu mentioned in comments, it's the same as counting $a$ such that $\gcd(a,n)=1$. Then, definitely $\gcd(a,n-a) =1$ and vice versa. So, the answer would be $\phi(n)$.
Where $\phi(n)$ is the Euler's Totient Function
