solving for positive integer solutions $x^2 = yz,$
$x + y + z = 61$
positive integral solutions.
On wolframalpha, i could find the solution, but not how to solve. 
I tried by taking a single equation 
$xz + x^2 +z^2 - 61z = 0$
Tried solving for both x & z separately but did not find a way.
Please help.
EDIT: This is a high school problem in numerical analysis. Diophantine Equations is not yet in course.

I wanted to see that a more algebraic approach looked like
$$\begin{align}
&\left[x^2=yz \quad | \quad z=61-x-y\right]\implies x^2=61y-xy-y^2 \\
&\qquad x^2+xy+\frac{y^2}{4}-\frac{y^2}{4}=61y-y^2 \\
&\qquad \left(x+\frac{y}{2}\right)^2=61y-\frac{3}{4}y^2 \\
&\qquad \left(2x+y\right)^2=244y-3y^2 \\ 
&\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\frac{122^2}{9}-\frac{244}{3}y+y^2 \\
&\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\left(\frac{122}{3}-y\right)^2 \\
&\qquad -3(2x+y)^2=-122^2+(122-3y)^2 \\
&\qquad (3y-122)^2+3(2x+y)^2=122^2=14884=2^2 \cdot 61^2 \implies\\
&\left[\alpha^2+3 \beta ^2=14884\right] \quad | \quad \alpha=3y-122  \ \ \land \ \ \beta= 2x+y
\end{align}$$
Whatever $\alpha,\beta$ that might solve the above
$$\begin{align}y=\frac{\alpha+122}{3} \ \ \land \ \ x&=\frac{\beta-y}{2} \\
&=\frac{\beta-\frac{\alpha+122}{3}}{2} \\
&=\frac{3\beta-\alpha -122}{6}\end{align}$$
$$ \underline{(\alpha_i, \beta_i) \to (x_i, y_i, z_i)} \\
(-74,56) \to (20,16,25) \\
 (-74,-56) \to (-36,16,81) \\
(-47,65) \to (20,25,16) \\
(-47,-65) \to (-45, 25,81) \\
(121,9) \to (-36,81,16) \\
(121,-9) \to (-45,81,25) \\
$$
$(\alpha, \beta)$ Solutions obtained through the help of Darios online equation solver: https://www.alpertron.com.ar/QUAD.HTM
