Graph of the equation

Question

The graph of the equation $$ x^2y^3=(2x+3y)^5$$ is same as $$x=-y$$ Why is this so?

I understand it satisfies the equation, but is there a way to derive it? Don't the other roots count? The graph

Edit:
Something I just realized. $$ dy/dx=y/x$$ for the equations $$x^my^n=(x+y)^{m+n}$$ we have the solution $$|y/x|=e^c$$ where c is a constant. Perhaps something similar is happening here?

Is there anything more general than this? Because it feels like a complex equation but turns out to be quite simple.

Answer 1

Since the equation is homogeneous, the equation defines a closed sub-variety of the projective space $\mathbb P_{\mathbb C}^1$, so it is a dimension 0 closed projective variety, which are just finitely many points. Thus in affine space the equation $x^2y^3 = (2x+3y)^5$ is just a union of finitely many lines of the form $y = ax$. Your question is then simplified to prove that $a = -1$ is the only real value for $a$. (another way to see this is via the dehomogenization $s = \frac yx$).

Let us prove that if $x \neq -y$, then $x^2y^3 \neq (2x+3y)^5$. Clearly if $x$ and $y$ have the same sign, then $|x^2y^3| < |(2x+3y)^5|$, and without loss of generality we can assume $y > 0$, $x < 0$ and $x+y > 0$. Taking $z = -x$ we then get $y > 0, z > 0$ and $y > z$, and

$$ z^2y^3 = (3y - 2z)^5. $$

By the arithmetic mean-geometric mean inequality,

$$ \sqrt[5]{z^2y^3} < \frac{z+z+y+y+y}{5} = \frac{2z+3y}{5} < \frac{2z+3y}{5} + \frac{12}{5}(y-z) = 3y-2z. $$

Answer 2

$$ \begin{aligned} (2x + 3y)^5 - x^2y^3 &= (2(x+y)+y)^5 - x^2y^3\\ &= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\ &\qquad - y^3(x^2 - y^2)\\ &= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\ &\qquad - y^3(x+y)(x+y-2y)\\ &= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 39(x+y)y^3 + 12y^4) \end{aligned} $$

Substitute $s = \frac{x+y}{y}$ you get an equation $32s^4 + 80s^3 + 80s^2 + 39s + 12 = 0$. Wolfram alpha tells you this equation has only complex solutions, so the second parenthesis in the first equation is positive unless $x=y=0$.