<s>[INST] System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Thank you for your time
 [/INST] I'm going to use $x=a_1, \ y=a_2, \ z=a_3$
$$xz - y^2=0 \tag A$$
$$x + z - 2y - 16 = 0 \tag B$$
$$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$
Substitute $xz = y^2$ from (A) into (C)
\begin{align}
   xz + 64x - y^2 - 16y - 64 &= 0 \\
   y^2 + 64x - y^2 - 16y - 64 &= 0 \\
   64x - 16y - 64 &= 0 \\
   y &= 4x - 4 \tag D
\end{align}
Substitute $y = (4x - 4)$ from (D) into (B)
\begin{align}
   x + z - 2y - 16 = 0 \\
      x + z - 2(4x - 4) - 16 = 0 \\
      -7x + z - 8 &= 0 \\
      z &= (7x + 8) \tag E
\end{align}
Substitute (D) and (E) into (A)
\begin{align}
   xz - y^2 &= 0 \\
   x(7x + 8) - (4x - 4)^2 &= 0 \\
   7x^2 + 8x - 16x^2 + 32x - 16 &= 0 \\
   -9x^2 + 40x - 16 &= 0 \\
   -(9x - 4)(x - 4) &= 0 \\
   x &\in \left\{ \dfrac 49, 4 \right\}
\end{align}
We get $$
   (a_1,a_2,a_3) = \left(\dfrac 49, -\dfrac{20}{9}, \dfrac{100}{9}  \right)
   \ \text{and} \ 
   (a_1,a_2,a_3) = (4, 12, 36)$$
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