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.

• It is easy to see that $x$ and $y$ could not be positive, I think. But real numbers, that's hard? – Sawarnik Feb 14 '14 at 14:13
• One place to start looking is where $x = y$. Then you have $$16^{x^2 + x} = \frac{1}{2} \implies x^2 + x = -\frac{1}{4}$$ so $x = y = -\frac{1}{2}$ is one solution (at least now we know that solutions exist). Also, we can easily see that both $x^2 + y$ and $y^2 + x$ must be negative. This limits the search a lot (they're both between $-1$ and $0$). – Arthur Feb 14 '14 at 14:15
• $16^{x(x-1)}+16^{y(y-1)}=16^{-(x+y)}$ – Lucian Feb 14 '14 at 14:30
• $16^{x^2+y}+16^{x+y^2} = 16^{x^2+y^2}(16^{y-y^2}+16^{x-x^2})$ – James S. Cook Feb 14 '14 at 14:32

$x^2 + y^2 + x + y = (x + 1/2)^2 + (y + 1/2)^2 - 1/2 \geq -1/2$ 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 $x=y=-1/2$

• Wow. beautiful method :) – MathMan Feb 14 '14 at 17:04

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\ol}{\overline{#1}}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert\, #1 \,\right\vert}$ $$\mbox{Let's}\quad 4^{x^{2} + y} = \cos\pars{\theta}\,,\quad 4^{y^{2} + x} = \sin\pars{\theta}\qquad\mbox{with}\qquad \theta\ \in\ \pars{0,{\pi \over 2}}$$

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

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

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

• So how do you find all solutions using that monster function? – Sawarnik Feb 20 '14 at 20:34

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 $A \ge 1$

$$16^{x^2 + y} + 16^{y^2 + x} = A$$

$x^2 + y = y^2 + x = \dfrac{ln{(\frac{A}{2})}}{ln{16}}$