Number of solutions to a polynomial

Question

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?

Answer 1

\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 $x-2y-1=4x-3y-4=0$.

Answer 2

$$\frac{a^2+b^2}{2}\ge ab$$ If $a=0, b=0$ then these two are equal and equal to $0$. Hence there are no other roots to the equation except when $$x-2y-2=0 \\ 4x-3y-4=0$$

Answer 3

Considering the previous answers, I could be totally wrong.

If the equation is $$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y\color{red}{+}4)=0$$ expand it as a polynomial of $x$ to get $$21 x^2- (39 y+34)x+(19 y^2+23 y+13)=0$$ Solve the quadratic to get $$x_\pm=\frac{1}{42} \left(39 y+34\pm\sqrt{-75 y^2+720 y+64}\right)$$ and notice that $$-75 y^2+720 y+64 \geq 0 \qquad \text{if}\qquad \frac{8}{15} \left(9-2 \sqrt{21}\right)\leq y \leq \frac{8}{15} \left(9+2 \sqrt{21}\right)$$ which seems to mean that there is an infinite number of solutions in the given range for $y$ which seems normal since the equation corresponds to a rotated ellipse.

Answer 4

Let $a=x-2y-1$ and $b=4x-3y+4$. Then we get $$a^2+ab+b^2 =0$$

Solving this quadratic equation on $a$ we get $D=-3b^2\leq 0$, so $b=0$ and then $a=0$. So we have to solve system if linear eqaution which is fairly simple...

Answer 5

Let $u=x-2y-1$ and $v=4x-3y-4$ so that the equation becomes $u^2+uv+v^2=0$. The discriminant of this conic is equal to $1^2-4=-3$, therefore it’s some kind of ellipse. The lack of constant and linear terms makes this ellipse degenerate—it’s a single point. An affine coordinate transformation can’t turn a degenerate ellipse into a nondegenerate one, so the original equation also has only one solution.