<s>[INST] Number of solutions to a polynomial I need to find the number of solution to the polynomial 

$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)=0$$

Clearly, $x=1$ and $y=0$ is a solution. How to prove it has no other solutions or more solutions?
 [/INST] \begin{align*}
&\;(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)\\
=&\;\frac{1}{2}(x-2y-1)^2+\frac{1}{2}(4x-3y-4)^2+\frac{1}{2}[(x-2y-1)+(4x-3y-4)]^2\\
=&\;0
\end{align*}
if and only if $x-2y-1=4x-3y-4=0$.
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