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Find all pairs (x, y) of real numbers such that

$$16^{x^2+y} + 16^{x+y^2}=1$$

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  • $\begingroup$ This is hard to parse. Do you mean $16(x^2+y)+16(x+y^2)=1$? $\endgroup$
    – lulu
    Jun 1, 2016 at 11:30
  • $\begingroup$ yes 16 is the base and the others are the exponents $\endgroup$
    – Pole_Star
    Jun 1, 2016 at 11:31
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    $\begingroup$ That is not what I wrote. so you mean $16^{x^2+y}+16^{x+y^2}=1$ $\endgroup$
    – lulu
    Jun 1, 2016 at 11:32
  • $\begingroup$ yeah the same.... $\endgroup$
    – Pole_Star
    Jun 1, 2016 at 11:33
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    $\begingroup$ This is hardly a problem about number theory. $\endgroup$
    – ajotatxe
    Jun 1, 2016 at 11:34

1 Answer 1

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A very nice problem!

By application of inequality $a+b\ge2\sqrt{ab}$ we have $$16^{x^2+y}+16^{x+y^2}\ge 2\sqrt{16^{x^2+y}\cdot 16^{x+y^2}}=2\cdot 4^{x^2+x+y^2+y}$$ Since $t^2+t\ge-\frac{1}{4}$ holds for all $t\in\mathbb{R}$, for all $x,y\in\mathbb{R}$ we have $$16^{x^2+y}+16^{x+y^2}\ge2\cdot 4^{x^2+x+y^2+y}\ge 2\cdot 4^{-\frac{1}{2}}=1$$ with equality only in case $x=y=-\frac{1}{2}$, so this is the only solution.

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    $\begingroup$ Over $\mathbb{C}$ we also have $x=(2i + 1)/2$, $y=( - 2i + 1)/2$ and vice versa. Then $x+y^2=x^2+y=-1/4$. $\endgroup$ Jun 1, 2016 at 11:44
  • $\begingroup$ thanks for the appreciation $\endgroup$
    – Pole_Star
    Jun 1, 2016 at 14:24

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