Find all triples $m$; $n$; $l$ of natural numbers such that 
$m + n = gcd(m; n)^2$; $m + l = gcd(m; l)^2$; $n + l = gcd(n; l)^2$:
[i]S. Tokarev[/i]
1. Let \( (m, n, l) = k \). We notice that \( \gcd(m, n, l)^2 \mid \gcd(m, n)^2 = m + n \). Let \( m = ka \), \( n = kb \), \( l = kc \). Then \( k \mid a + b \), \( k \mid b + c \), \( k \mid c + a \). Clearly, \( k \mid 2a + 2b + 2c \).

2. If \( k \) is odd, then \( k \mid a + b + c \). Since \( k \mid a + b \), \( k \mid b + c \), and \( k \mid c + a \), we have \( k \mid a \), \( k \mid b \), and \( k \mid c \). But since we took out their gcd already, \( \gcd(a, b, c) = 1 \), so we need \( k = 1 \).

3. Suppose \( k \) is even, so let \( k = 2x \). Then \( x \mid a + b + c \), \( x \mid a + b \), \( x \mid b + c \), and \( x \mid c + a \). By the same reasoning, \( x \mid a \), \( x \mid b \), and \( x \mid c \). Thus, \( x = 1 \) and \( k = 2 \).

4. Now return to the case \( k = 1 \). We see the existence of constants such that \( m = vwa \), \( n = uwb \), \( l = uvc \) where \( \gcd(a, b) = \gcd(b, c) = \gcd(c, a) = 1 \). Then \( va + ub = w \), \( wa + cu = v \), \( wb + vc = u \). Summing these gives \( u + v + w = u(b + c) + v(c + a) + w(a + b) \). Since \( m, n, l \) were natural, we see that \( a, b, c \) are positive, so obviously the LHS < RHS.

5. Take \( k = 2 \) and we get that \( m = 2vwa \), \( n = 2uwb \), \( l = 2uvc \) and \( va + ub = 2w \), \( wa + cu = 2v \), \( wb + vc = 2u \). So \( 2u + 2v + 2w = u(b + c) + v(c + a) + w(a + b) \geq 2u + 2v + 2w \) with equality when \( a = b = c = 1 \). Thus we need \( a = b = c = 1 \) and then \( v + u = 2w \), \( w + u = 2v \). The third equation is redundant. However, here we obtain \( 2w + 2u = 4v \) so \( 3u + v = 4v \) and \( u = v \). Similarly, we get \( v = w \) so \( u = v = w \) and then we conclude they must all equal 1 as \( \gcd(u, v, w) = 1 \). So we see the only solution is \( (m, n, l) = (2, 2, 2) \).

The final answer is \( \boxed{ (2, 2, 2) } \).