$m,n,r\in(0;0;0) and (2;2;2)$

Let $m\ge n\ge r\ge 3$ then $n^2+3n+3m\ge mn+nr+mr\ge=2(m+n+r)\ge 2m+2n+m+n=3m+3n$

Suppose $m\ge n\ge r\ge 3$ then $3m+3n=2m+2n+m+n\ge 2m+2n+2r=mn+nr+mr\ge 3m+3n+n^2$

You are still missing $(1,2,4),(1,4,2),(2,1,4),(2,4,1),(4,1,2),(4,2,1)$. Together with $(2,2,2)$, there are $7$ solutions in positive integers (all your other solutions contain $0$, and that is not a positive integer). If you also bothered to read point (2), that would have warned you you should get at least $3\cdot 2 + 1 = 7$ solutions in positive integers (the case $k=2$). For point (2), if $m,n,r > k$ it is clear there are no solutions. Assume therefore $1\leq \min\{m,n,r\} \leq k$. Notice $(k,k,k)$ is always a solution. Now, write it as $(m - (k-r))(n-(k-r)) = k^2 - kr + r^2$ for $1\leq r\leq k$. Then $(k-r+1, k^2-kr+r^2 + k - r + 1, r)$ and $(k^2-kr+r^2 + k - r + 1,k-r+1, r)$ are solutions. Now we have to check if, when running $r$ in $\{1,2,\ldots,k\}$, and when permuting $m,n,r$ we don't get duplicate solutions (which we will); but I will not get into these details.

I'm very verryyyy sorry. I had a mistake again and forgot these solutions Thanks a lot mavropmevma!

mavropnevma wrote: You are still missing $(1,2,4),(1,4,2),(2,1,4),(2,4,1),(4,1,2),(4,2,1)$. Together with $(2,2,2)$, there are $7$ solutions in positive integers (all your other solutions contain $0$, and that is not a positive integer). If you also bothered to read point (2), that would have warned you you should get at least $3\cdot 2 + 1 = 7$ solutions in positive integers (the case $k=2$). For point (2), if $m,n,r > k$ it is clear there are no solutions. Assume therefore $1\leq \min\{m,n,r\} \leq k$. Notice $(k,k,k)$ is always a solution. Now, write it as $(m - (k-r))(n-(k-r)) = k^2 - kr + r^2$ for $1\leq r\leq k$. Then $(k-r+1, k^2-kr+r^2 + k - r + 1, r)$ and $(k^2-kr+r^2 + k - r + 1,k-r+1, r)$ are solutions. Now we have to check if, when running $r$ in $\{1,2,\ldots,k\}$, and when permuting $m,n,r$ we don't get duplicate solutions (which we will); but I will not get into these details. Mavropnevma, how can we check when running $r$ in $\{1,2,\ldots,k\}$, and when permuting $m,n,r$ we don't get duplicate solutions? Thanks a lot in advance.