<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] Subtract the first row from the last row of the system: then you get
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1  -16 a_2 -64 &=0
\end{split}
$$
Then multiply by 8 the second row and subtract it again from the third row:
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1  -16 a_2 -64 &=0
\end{split}\implies
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
56a_1 -8a_3+64 &=0
\end{split}
$$
Then you can proceed by substitution and do the following steps
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
a_1 &=\frac{1}{7}(a_3-8)
\end{split}\implies
\begin{split}
\frac{1}{7}a_3^2-\frac{8}{7}a_3-a_2^2 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1&=\frac{1}{7}(a_3-8)
\end{split}
$$
and finally we have 
$$
\begin{split}
9a_3^2-424a_3-3600 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1  &=\frac{1}{7}(a_3-8)
\end{split}
$$
where the first equation is a quadratic equation respect to the single $a_3$.
In sum, a nice way to proceed for such system is to see if you can subtract a multiple of one row from another, in order to simplify the structure and possibly arrive to a system where all the equations are linear except one, which however contains only one of the variables.
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