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.