<s>[INST] Find all real solutions for $16^{x^2 + y} + 16^{x + y^2} = 1$ Find all $x, y \in \mathbb{R}$ such that:
$$16^{x^2 + y} + 16^{x + y^2} = 1$$
The first obvious approach was to take the log base $16$ of both sides:
$$\log_{16}(16^{x^2 + y} + 16^{x + y^2}) = 0$$
manipulating did not give any useful result. The next thing I tried was getting some bounds on $x$ and $y$:
If $x, y \geq 0$, 
$$16^{x^2 + y} + 16^{x + y^2} \geq 2$$
So, $x, y \le 0$. Trying to obtain a lower bound was not fruitful. 
Also, in general, I have a lot of difficulty solving such problems which require all solutions to a certain equation. 
Whatever I do is almost always contrary to what the actual solution is and the solution itself involves some bizarre counter-intuitive manipulations or methods. Some tips on how to approach such problems will be helpful for me. Thanks. 
 [/INST] Using $$\bf{A.M\geq G.M}$$,
$\displaystyle \frac{16^{x^2+y}+16^{x+y^2}}{2}\geq \left(16^{x^2+y}\cdot 16^{x+y^2}\right)^{\frac{1}{2}}$
$\displaystyle 16^{x^2+y}+16^{x+y^2}\geq 2\cdot \left\{2^{4\left(x^2+y+x+y^2\right)}\right\}^{\frac{1}{2}} =  2\cdot \left\{2^{(2x+1)^2+(2y+1)^2-2}\right\}^{\frac{1}{2}}\geq 1$
and equality hold when $16^{x^2+y} = 16^{x+y^2}\Rightarrow x^2+y = x+y^2\Rightarrow \displaystyle x=y = -\frac{1}{2}$
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