<s>[INST] How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$ 
show that there exsit infinitely many  postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$

May try it is clear $(x,y,z)=(1,1,1)$ is one  solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
 [/INST] If $(1,y,z)$ is a solution then $(1,z,3z-y+1)$ is also a solution because \begin{align*} &\quad \Big( 1 + z + (3z - y + 1) \Big)^2 + 2 \Big( 1 + z + 3z - y + 1 \Big) - 5 \Big( z + z(3z - y + 1) + 3z - y + 1 \Big) \\ &= (1 + y + z)^2 + 2(1 + y + z) - 5(y + yz + z). \end{align*} You can use this to generate the infinite family
$(1,1,1)$, $(1,1,3)$, $(1,3,9)$, $(1,9,25)$, $(1,25,67), ...$ of solutions.
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