<s>[INST] 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.
 [/INST] $x^2 + y^2 + x + y = (x + 1/2)^2 + (y + 1/2)^2 - 1/2 \geq -1/2$ 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 $x=y=-1/2$
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