Question

Find all pairs $(x,y)$ of real numbers such that ${16}^{x^2+y}+{16}^{x+y^2}=1$.

Answer 1

We have, ${16}^{x^2+y}+{16}^{x+y^2}=1$
$x^2+y+x+y^2+\frac{1}{2}={(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2\ge 0$
This shows that $x^2+y+x+y^2\ge -\frac{1}{2}$
By AM-GM inequality,
$1={16}^{x^2+y}+{16}^{x+y^2}\ge 2{\left({16}^{x^2+y}{.16}^{x+y^2}\right)}^{\frac{1}{2}}$
$2{\left({16}^{x^2+y+x+y^2}\right)}^{\frac{1}{2}}\le 2(16{)}^{-\frac{1}{4}}=1$ .
This equality holds every where
$\therefore {(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2=0$
$\therefore (x,y)=(\frac{1}{2},\frac{1}{2})$