14
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ It is easy to see that $x$ and $y$ could not be positive, I think. But real numbers, that's hard? $\endgroup$ – Sawarnik Feb 14 '14 at 14:13
  • 7
    $\begingroup$ 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 $0$). $\endgroup$ – Arthur Feb 14 '14 at 14:15
  • $\begingroup$ $16^{x(x-1)}+16^{y(y-1)}=16^{-(x+y)}$ $\endgroup$ – Lucian Feb 14 '14 at 14:30
  • $\begingroup$ $16^{x^2+y}+16^{x+y^2} = 16^{x^2+y^2}(16^{y-y^2}+16^{x-x^2})$ $\endgroup$ – James S. Cook Feb 14 '14 at 14:32
18
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Wow. beautiful method :) $\endgroup$ – MathMan Feb 14 '14 at 17:04
0
$\begingroup$

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

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

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

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

$\endgroup$
  • $\begingroup$ So how do you find all solutions using that monster function? $\endgroup$ – Sawarnik Feb 20 '14 at 20:34
0
$\begingroup$

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 $A \ge 1$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.