Let $n\in\mathbb{Z}_+$ then $n=\sum_{d|n} \phi(d)$, where $\sum_{d|n} $ denotes the sum over all of the divisors of $n$ and $\phi(d)$ is the Euler phi function which gives the number of integers less than $d$ which are coprime to $d$.
In order to prove this, define set $$ A(d)=\{m\in\mathbb{Z}|1\leq m\leq n\quad\&\quad gcd(m,n)=d\}\\ gcd(m,n)=d: d|m \Leftrightarrow m=ds\quad\&\quad d|n \Leftrightarrow n=dt\text{ for some }s,t\in\mathbb{Z}\\ \text{Bezout's identity : } d=mx+ny \text{ for some }x,y\in\mathbb{Z}\\ \implies d=dsx+dty\implies 1=sx+ty\\ \implies gcd(s,t)=gcd(m/d,n/d)=1 $$ Since $A(d)$ is all the numbers less than or equal to $n$ whose gcd with $n$ is $d$, implies the number of integers in $A(d)$, i.e., $|A(d)|$ equals the number of positive inegers less than or equal to $n/d$ that are coprime to $n/d$. Therefore, $$ |A(d)|=\phi(n/d) $$
This much is clear. But, in order to finish the proof we need to prove that $A(d)$ are pairwise disjoint, how does one show this ?
My Attempt
Each such sets can be restated as, $A(n/d)=\{m/d\in\mathbb{Z}|1\leq m/d\leq n/d\quad\&\quad gcd(m/d,n/d)=1\}$
$d_1,\cdots,d_r$ are divisors of $n$, therefore $n/d_1,\cdots,n/d_r$ are divisors of $n$ in a different same order.
So we can restate the sets as $A(d)=\{m\in\mathbb{Z}|1\leq m\leq n\quad\&\quad gcd(m,n)=1\}$ and we have show them as pairwise disjoint.