Find all such triples of natural numbers $m, n$ and $l$, that $m+n=(\text{GCD}(m, n))^{2}, m+l=(\text{GCD}(m, l))^{2}, n+l=$ (GCD $(n, l))^{2}$.
Let $d=$ GCD $(m, n, l)$. Let $m=d m_{1}, n=d n_{1}, l=d l_{1}$. Then $d\left(m_{1}+n_{1}\right)=m+n=d^{2} d_{m n}^{2}$, where $d_{m n}=$ GCD $\left(m_{1}, n_{1}\right)$; from which $m_{1}+n_{1}=d d_{m n}^{2}$. Adding this equality to two similar ones, we get $2\left(m_{1}+n_{1}+l_{1}\right)=d\left(d_{m n}^{2}+d_{n l}^{2}+d_{m l}^{2}\right)$

We will show that $d$ is coprime with the sum $m_{1}+n_{1}+l_{1}$. Indeed, if $d$ and this sum have a common divisor $\delta>1$, then it will be a common divisor of all numbers $m_{1}, n_{1}$ and $l_{1}$ (since the sum of any two of them is divisible by $d$). But then $d \delta$ - is a common divisor of the numbers $m, n$ and $l$, which contradicts the definition of the number $d$.

Therefore, $d$ is a divisor of the number 2.

Note that the numbers $d_{m n}, d_{n l}, d_{m l}$ are pairwise coprime (otherwise, the numbers $m_{1}, n_{1}, l_{1}$ would have a common divisor not equal to 1). Therefore, $m_{1}=d_{m n} d_{m l} m_{2}$,

$n_{1}=d_{m n} d_{n l} n_{2}, l_{1}=d_{n l} d_{m l} l_{2}$, where $m_{2}, n_{2}, l_{2}$ are natural numbers. In such notation, the first of the original equations takes the form $d_{m l} m_{2}+d_{n l} n_{2}=d d_{m n}$.

Without loss of generality, we can assume that the number $d_{m n}$ is the smallest of the numbers $d_{m n}, d_{m l}$ and $d_{n l}$. We have: $d_{m l} m_{2}+$ $d_{n l} n_{2} \geq d_{m l}+d_{n l} \geq 2 d_{m n} \geq d d_{m n}$ (since

$d \leq 2$ ). Thus, all inequalities are actually equalities, from which $m_{2}=n_{2}=1, d=2$ and $d_{m l}=d_{m n}=d_{n l}$. But the numbers $d_{m l}, d_{m n}, d_{n l}$ are pairwise coprime, therefore, they are equal to 1, and we have found the only solution $m=n=l=2$.

## Answer

$(2,2,2)$.