2022/019) Find real x and y such that $16^{x^2+y} + 16^{y^2 + x} =1$

 We have both $16^{x^2+ y}$ and $16^{y^2+x}$ positive.

So we can apply AM GM inquality gettting

$\frac{16^{x^2+y} + 16^{y^2+x}}{2}\ge (16^{x^2+y} 16^{y^2+x})^{\frac{1}{2}}$

or $16^{x^2+y} + 16^{y^2+x}\ge  2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

we are given LHS = 1 so 

$ 1 \ge 2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

or $ 1 \ge 16^{\frac{1}{4}}  (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

or $ 1 \ge  (16^{x^2+y+ y^2+x+ \frac{1}{2}})^{\frac{1}{2}}$

or $x^2 + y + y^2 + x + \frac{1}{2} <=0$ 

Now we combine line terms and complete square to get 

$(x^2 + x + \frac{1}{4}) + (y ^2 + y + \frac{1}{4}) <= 0$

or $(x+\frac{1}{2})^2 + (y + \frac{1}{2})^2 <=0$

it is sum of 2 squares so cannot be be -ve so

both terms and sum has to be zero givng $x = y = \frac{-1}{2}$