Finding the real solutions to $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ We have , $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ , then we have to find all the real values of $x$ and $y$.I tried this question  but i am not able to proceed because I am not able to simplify this expression to an extent that it could be solved.
 A: $x^2 + y^2 + x + y = (x + 1/2)^2 + (y + 1/2)^2 - 1/2 \geq -1/2$ and equality occurs only when $x = y = - 1/2$.
Using AM-GM inequality $16^{x^2 + y} + 16^{y^2 + x} \geq 2\cdot\sqrt{16^{x^2+y^2+x+y}} \geq 2\cdot16^{-1/4} = 1$ and equality occurs only when $x=y=-1/2$
A: $\newcommand{\+}{^{\dagger}}%
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$$
\mbox{Let's}\quad 4^{x^{2} + y} = \cos\pars{\theta}\,,\quad
4^{y^{2} + x} = \sin\pars{\theta}\qquad\mbox{with}\qquad \theta\ \in\ \pars{0,{\pi \over 2}}
$$

$$
x^{2} + y ={\ln\pars{\cos\pars{\theta}} \over 2\ln\pars{2}}\,,\qquad
y^{2} + x ={\ln\pars{\sin\pars{\theta}} \over 2\ln\pars{2}}
$$

$$
\bracks{{\ln\pars{\cos\pars{\theta}} \over 2\ln\pars{2}} - x^{2}}^{2} + x ={\ln\pars{\sin\pars{\theta}} \over 2\ln\pars{2}}
$$

$$
x^{4} - {\ln\pars{\cos\pars{\theta}} \over \ln\pars{2}}\,x^{2} + x
+
\bracks{{\ln^{2}\pars{\cos\pars{\theta}} \over 4\ln^{2}\pars{2}}
- {\ln\pars{\sin\pars{\theta}} \over 2\ln\pars{2}}} = 0\,,\qquad
\theta\ \in\ \pars{0,{\pi \over 2}}
$$
  which express $x$ ( parametrically ) as a function of $\theta$.

A: This is not exactly the original problem ,  but it is related to a procedure suggested by Arthur. It looks like we can always find real number solutions $(x , y)$ for $A \ge 1$
$$ 16^{x^2 + y} + 16^{y^2 + x} = A $$
$x^2 + y = y^2 + x = \dfrac{ln{(\frac{A}{2})}}{ln{16}}$ 
