# Find $y$ if $x^{x+y}=y^n$ and $y^{x+y}=x^{2n}y^n$, where $x,y,n>0$

If $$x,y>0$$ satisfying the system of equations $$x^{x+y}=y^n$$ and $$y^{x+y}=x^{2n}y^n$$, where $$n>0$$ then prove that $$y=\dfrac{1+4n-\sqrt{1+8n}}{2}$$

$$(xy)^{x+y}=(xy)^{2n}\implies x+y=2n\\ x^{2n}=y^n\implies x^2=y\\ x^{2}=2n-x\implies x^2+x-2n=0\\ x=\frac{-1\pm\sqrt{1+8n}}{2}\implies x=\frac{-1+\sqrt{1+8n}}{2}\\ y=x^2=\frac{2+8n-2\sqrt{1+8n}}{4}=\frac{1+4n-\sqrt{1+8n}}{2}$$ Fine, but if I do the opposite $$y=x^2=(2n-y)^2=4n^2+y^2-4ny\\ y^2-(4n+1)y+4n^2=0\\ y=\frac{1+4n\pm\sqrt{1+8n}}{2}$$ In the second approach how do I eliminate the other case ?

Attempt $$y=x^2=\frac{1+4n\pm\sqrt{1+8n}}{2}=\frac{2+8n\pm2\sqrt{1+8n}}{4}=\bigg[\frac{1\pm\sqrt{1+8n}}{2}\bigg]^2\\ \implies x=\bigg|\frac{1\pm\sqrt{1+8n}}{2}\bigg|$$ $$y=\frac{1+4n+\sqrt{1+8n}}{2}\implies x=\frac{1+\sqrt{1+8n}}{2}$$ $$y=\frac{1+4n-\sqrt{1+8n}}{2}\implies x=\frac{-1+\sqrt{1+8n}}{2}$$ Still getting two cases ?

Note that $$x+y=2n$$, and since $$x,y,n$$ are positive, it must be that $$x=2n-y>0$$, that is, $$y<2n$$. However, the superfluous solution does not satisfy this.