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
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
 A: Lemma($\color{Green}{\text{Vieta's formula}}$):
Let $\alpha_1$ 
be the root of the quadratic polynomial equation 
$aY^2+bY+c=0$;
then we have: $\alpha_2= \color{Blue}{\dfrac{-b}{a}}-\alpha_1$. 
Proof: 
Only notice that 
$\alpha_1 + \alpha_2 = \dfrac{-b}{a}$. 


$ 
\color{Purple} 
{ 
\text{Let's to look at one of the} 
\ \ 
x, y, z 
\ \ 
\text{as the} 
} $ 
$\color{Red}{\text{variable}}$ 
$ 
\color{Purple} 
{  
\text  {and to look at the others as} 
} $ 
constants.  

For example let's to look at $\color{Red}{y}$ as an indeterminate, 
and to look at $x,z$ as constants;
as @user399601 has been done.  
$$(x+\color{Red}{y}+z)^2+2(x+\color{Red}{y}+z)=5(x\color{Red}{y}+\color{Red}{y}z+zx) 
\Longrightarrow 
\\ 
\Bigg[ 
\color{Red}{y^2} 
+ 
\big(2(x+z)\big) 
\color{Red}{y} 
+ 
(x+z)^2 
\Bigg] 
+ 
\Bigg[ 
2 
\color{Red}{y} 
+(x+z) 
\Bigg] = 
\Bigg[ 
5(x+z) 
\color{Red}{y} 
+ 5zx 
\Bigg] 
\Longrightarrow 
\\ 
\color{Red}{y^2} 
+ 
\Big( 
2(x+z) 
+ 
2 
-5(x+z)
\Big) 
\color{Red}{y} 
+ 
\Big( 
(x+z)^2 
+ 
(x+z) 
-5zx 
\Big) 
=0  
\ \ \ \ \ \ \ \ \ \ \ 
\Longrightarrow 
\\ 
\color{Red}{y^2} 
+ 
\Big( 
\color{Blue}{ 
2 
-3(x+z)
} 
\Big) 
\color{Red}{y} 
+ 
\Big( 
(x+z)^2 
+ 
(x+z) 
-5zx 
\Big) 
=0  
\ \ \ \ \ \ \ 
\color{Green}{\star\star\star\star} 
$$


Suppose that 
$\color{Red}{y}$ 
satisfies the polynomial equation 
$\color{Green}{\star\star\star\star}$ 
;
then by 
$\color{Green}{\text{Vieta's formula}}$ 
; 
we can see that 
: 
$ 
\Big( 
\color{Blue}{ 
3(x+z) 
-2 
} 
- \color{Red}{y} 
\Big) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
. 
So we proved that:

If 
$(x,\color{Red}{y},z)$ satisfies 
$\color{Green}{\star\star\star\star}$ ; 
then 
$ 
( 
x 
, 
\color{Blue}{ 
3(x+z) 
-2 
} 
- \color{Red}{y} 
, 
z 
) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
.  
[ 
More specially if we let $x=1$ we have the following:
If 
$(1,\color{Red}{y},z)$ satisfies 
$\color{Green}{\star\star\star\star}$ ; 
then 
$ 
( 
x 
, 
\color{Blue}{ 
3z 
+ 
3 
-2 
} 
- \color{Red}{y} 
, 
z 
) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
. 
] 




$ 
\color{Purple} 
{  \text  {This method is called}}$ 
$ 
\color{Green} 
{  \text  {vieta-jumping}}$ 
For more information you can see here: 
https://math.stackexchange.com/questions/tagged/vieta-jumping
https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping
A: This gives exactly the sort of jumping that creates the Markov Tree 
Since the order of $x,y,z$ does not matter, it is traditional to order $1 \leq x \leq y \leq z$ to save space. Then, we get two jumps that go to the next bigger layer in the tree. The slower growth is ( I am ordering the outcome as well)
$$ (x,y,z) \mapsto (x,z, 3x+3z - y - 2).  $$ The faster growth is
$$ (x,y,z) \mapsto (y,z, 3y+3z - x - 2).  $$
This diagram is closer to the view in the wikipedia article

The first few layers are

How do we find the relevant formulas for the next two leaves, coming out of an existing leaf $(x,y,z)?$  This part is called Vieta Jumping. We have an (positive) integer solution to 
$$ x^2 + (2 - 3y - 3z)x + \; \mbox{stuff} = 0. $$
If the $x$ value we have and the other solution to the quadratic is called $x',$ we have
$$ x + x' = 3y + 3z - 2, $$ so that
$$  x' = 3y+3z-x-2, $$ in ascending order we get $(y,z,x').$
If we are going to flip the $y $ value instead, we have
$$ y^2 + (2 - 3x - 3z)y + \; \mbox{stuff} = 0. $$
If the $y$ value we have and the other solution to the quadratic is called $y',$ we have
$$ y + y' = 3x + 3z - 2, $$ so that
$$  y' = 3x+3z-y-2, $$ in ascending order we get $(x,z,y').$
The best discussion of this that I know is a 1907 article by Hurwitz in German. In preparing my article in this field with Kaplansky, I relied on  Cusick, Thomas; Flahive, Mari (1989). The Markoff and Lagrange spectra. 
A: 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.
A: For the equation.
$$(x+y+z)^2+2(x+y+z)=5(xy+xz+yz)$$
It is possible to reduce the parameterization of the solutions to some equivalent to the Pell equation.
It has the form.
$$x=3a^2-(b+c)a+b^2-3bc+c^2$$
$$y=a^2-(b+3c)a+3b^2-bc+c^2$$
$$z=a^2-(3b+c)a+b^2-bc+3c^3$$
These parameters can be recorded through the solution of the equation Pell.
$$p^2-5(2k^2+t^2)s^2=-1$$
$$a=ks$$
$$b=p-(3k-t)s$$
$$c=p-(3k+t)s$$ 
A: Here's a way to more constructively get the answer user399601 provided:
Note that since $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$, the equation 
\begin{equation}\label{eqn:constraint}\tag{1}
(x + y + z)^2 + 2(x + y + z) = 5(xy + yz + xz)
\end{equation}
is equivalent to
\begin{align*}
x^2 + y^2 + z^2 + 2(x + y + z) &= (x + 1)^2 + (y + 1)^2 + (z + 1)^2 - 3\\
&= 3xy + 3yz + 3xz
\end{align*}
Now take $z = 1$ to get
\begin{equation*}
(x + 1)^2 + (y + 1)^3 + 1 = 3(x + 1)(y + 1)
\end{equation*}
Making the substitutions $u = x+1$ and $v = y + 1$, the equation becomes
\begin{equation}\label{eqn:simplified}\tag{2}
u^2 + v^2 + 1 = 3uv
\end{equation}
and
\begin{equation*}
(u - v)^2 + 1 = uv
\end{equation*}
Now suppose there's some function $c:\mathbb{Z}^2 \to \mathbb{Z}$ such that $(u + c, v)$ is a solution whenever $(u, v)$ is a solution.  Then
\begin{align*}
(u - v + c)^2 + 1 &= (u + c)v\\
(u - v)^2 + 2c(u-v) + c^2 + 1 &= uv + cv \\
c^2 + 2(u - v)c &= cv
\end{align*}
So $$c(u, v) = 2(v - u) + v = 3v - 2u$$
Now note that if $(u, v)$ solves equation \eqref{eqn:simplified}, then $(u-1, v-1, 1)$ solves equation \eqref{eqn:constraint}.  Thus the solution $(3v - u, v)$ maps to $(3y - x + 1, y, 1)$.
