The problem doesn't actually require $ m,n,l$ to be different. Anyway, this doesn't change anything. I will use $ (m,n)$ for $ \gcd(m,n)$ in this solution. Let $ (m,n) = a$, $ m = am'$. From $ m + n = a^2$, we get $ n = a^2 - am'$. In a similar way, let $ (m,l) = b$, $ m = bm''$ and $ l = b^2 - bm''$. Because $ (m',a - m') = 1$ and $ (m'',b - m'') = 1$ we have $ (a,m') = (b,m'') = 1$. From $ am' = bm''$ we deduce the existence of $ 4$ positive integers $ x,y,z,t$ pairwise coprime, such that $ a = xy$, $ m' = zt$, $ b = xz$, $ m'' = yt$. This is really a well-known fact; anyway, for showing it, start by taking $ \dfrac ab = \dfrac{m''}{m'} = \dfrac yz$, with $ (y,z) = 1$. Then $ n = a^2 - am' = x^2y^2 - xyzt$ and $ l = b^2 - bm'' = x^2z^2 - xyzt$. We have $ (n,l) = x(xy^2 - yzt,xz^2 - yzt) = x(xy - zt,xz - yt)$, because $ (y,xz^2) = (z,xy^2) = 1$. Then $ (n,l)^2 = n + l$ rewrites as $ x^2(xy - zt,xz - yt)^2 = x^2(y^2 + z^2) - 2xyzt$. This means $ 2xyzt$ is divisible by $ x^2$. But $ (x,yzt) = 1$, from where $ x|2$. Case 1 $ x = 2$. Because we're working with positive integers, $ xy - zt$ and $ xz - yt$ are positive, so $ 2y > zt$ and $ 2z > yt$, from where $ t = 1$. Then $ y^2 + z^2 - yz = (2y - z,2z - y)^2$. Suppose $ p^k||(2y - z,2z - y)$, where $ p$ is a prime number. Then $ p^k|4y - 2z + 2z - y$, or $ p^k|3y$. Similarly $ p^k|3z$. Since $ (y,z) = 1$, $ p = 3$ and $ k = 1$ or $ p$ doesn't exist at all. This means $ y^2 + z^2 - yz\in\{1,9\}$. The equation $ y^2 + z^2 - yz = 1$ has the only solution $ y = z = 1$. This case leads to $ m = n = l = 2$. Let's analyze $ y^2 + z^2 - yz = 9$. WLOG, let $ y\ge z > 0$ and let $ y = z + k$. Then the last equation becomes $ z^2 + zk + k^2 = 9$. The only solutions are $ z = 3$ and $ k = 0$. But then $ y = z = 3$ and $ (y,z)\neq1$. Case 2 $ x = 1$. Again, because we're working with positive integers, we must have $ xy > zt$ and $ xz > yt$, or $ y > zt$ and $ z > yt$, from where $ 1 > t$, impossible. The only such numbers are $ m = n = l = 2$.

Also see page 99 (or 100) \ problem 38.\ of this book: http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf

Another method. Suppose that l is smaller than m and n. Then 2l < l+m = gcd(l,m)^2. Similarly 2l < l + n = gcd(l,n)^2. Multiply together and square rooting, we get that 2l < gcd(l,m)gcd(l,n). It is well know that gcd(u,v)lcm(u,v)=uv for all integers u,v. Thus gcd(l,m)gcd(l,n)=gcd(gcd(l,m),gcd(l,n))lcm(gcd(l,m),gcd(l,n)). The first term in this product is gcd(l,m,n). The second is a factor of l. Thus gcd(l,m)gcd(l,n) is a factor of gcd(l,m,n)l. Since 2l < gcd(l,m)gcd(l,n) we see that 2 < gcd(l,m,n). Now let z = gcd(l,m,n). z^2 is a factor of (gcd(l,m)^2)=l+m. z^2 is also a factor of m+n and l+n. Thus z^2 is a factor of (l+m)+(l+n)-(m+n)=2l. Similarly z^2 is a factor of 2m and 2n. Thus z^2 is a factor of gcd(2m,2l,2n)=2z. Thus z is a factor of 2. CONTRADICTION

$ gcd(l,m)^t=l+m, gcd(m,n)^t=m+n, gcd(n,l)^t=n+l,t>2$ has no solutions in N.

Vinxenz wrote: $ gcd(l,m)^t=l+m, gcd(m,n)^t=m+n, gcd(n,l)^t=n+l,t>2$ has no solutions in N. Can somebody solve the problem above?

Vinxenz wrote: $ gcd(l,m)^t=l+m, gcd(m,n)^t=m+n, gcd(n,l)^t=n+l,t>2$ has no solutions in N. Letting $l=gx,m=gy,n=gz$ with $g=\gcd(l,m,n)$ so that $\gcd(x,y,z)=1$, we have \[g^t|\gcd(l,m)^t=l+m=g(x+y)\implies g^{t-1}|x+y,\]etc. So $g^{t-1}|(x+y)+(x+z)-(y+z)=2x$, etc., whence $g^{t-1}|2$ and $g=1$. Thus letting $u=\gcd(y,z),v=\gcd(z,x),w=\gcd(x,y)$, we have $x=x'vw$, $y=y'wu$, and $z=z'uv$, where the triples $(x',y',z')$ and $(u,v,w)$ contain pairwise relatively prime elements, with $u^{t-1}=y'w+z'v$, $v^{t-1}=z'u+x'w$, and $w^{t-1}=x'v+y'u$. Now assume for the sake of contradiction that $p|\gcd(x',u)$ for a prime $p$. Then we find from our equations that $p|v,w$, a contradiction. Similarly, if some prime $q|\gcd(x',v)$, the fact that $\gcd(x',y')=\gcd(x',z')=1$ gives us $q|u,w$, another contradiction. We conclude that $x',y',z',u,v,w$ are all pairwise relatively prime. Note that \[(x',y',z')=\left(\frac{v^t+w^t-u^t}{2vw},\frac{w^t+u^t-v^t}{2wu},\frac{u^t+v^t-w^t}{2uv}\right).\]With some computing power, we find that $t=6$ actually has some solutions (there don't seem to be any under 9000 for $3\le t\le5$, although $t=4$ seems to have many examples where one of $x',y',z'$ is an integer and the other two are half-integers); the smallest example is \begin{align*} u &= 65 = 5\cdot13\\ v &= 72 = 2^3\cdot3^2\\ w &= 73\\ x' &= 20474639 = 2029\cdot10091\\ y' &= 9213809 = 11\cdot837619\\ z' &= 6773369 = 109\cdot62141\\ l &= gx = x'vw = 2^3\cdot3^2\cdot73\cdot2029\cdot10091=107614702584\\ m &= gy = y'wu = 5\cdot11\cdot13\cdot73\cdot837619=43719523705\\ n &= gz = z'uv = 2^3\cdot3^2\cdot5\cdot13\cdot109\cdot62141=31699366920. \end{align*} Whoops I just wasted a few hours on this.