<s>[INST] Equation has exactly one real solution I am trying to show that the equation
$$
4x^2y^4+12x^2y^2+4x^2+4xy^2+4x+1=0
$$
has exactly one real solution $x,y$ and to determine it.
The first observation is: If $y=0$, we are left with
$$
4x^2+4x+1=0
$$
which is solved by $x=-1/2$.
Thus a real solution is given by
$$
x=-\frac{1}{2}, y=0
$$
and it remains to show that this is the only real solution.
This is where I am stuck...
 [/INST] I do not know if it helps, but you can rearrange the proposed equation as follows:
\begin{align*}
4x^{2}y^{4} + 12x^{2}y^{2} + 4x^{2} + 4xy^{2} + 4x + 1 = 0 & \Longleftrightarrow (4x^{2}y^{4} + 4xy^{2} + 1) + (12x^{2}y^{2} + 4x^{2} + 4x) = 0\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (4x^{2} + 4x + 1) = 1\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (2x + 1)^{2} = 1
\end{align*}
</s>