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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.

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  • $\begingroup$ You can easily find all solutions with $x=y$. I doubt that's all of them, though. $\endgroup$ Nov 17, 2013 at 7:07
  • $\begingroup$ Since$16^x\ge0$,we have for all solutions $x+y^2\le0;y+x^2\le0.$ $\endgroup$
    – asatzhh
    Nov 17, 2013 at 7:13

2 Answers 2

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By the AM-GM inequality, (since $16^x>0$) \begin{align}16^{x^2+y} + 16^{y^2+x} &\geq 2\times4^{x^2+x}\times4^{y^2+y}\\ &=4^{x^2+x+1/4}\times4^{y^2+y+1/4}\\ &=4^{(x+1/2)^2}\times4^{(y+1/2)^2}\\ &\geq4^0\times4^0\\ &=1 \end{align}

The second inequality comes from the sum of squares being always non-negative.

Hence for overall equality, both inequalities must be equalities. Equality for sum of squares being greater than $0$ is if both squares are zero, i.e. $x, y = -1/2$ here.

Fortunately this also gives equality in AM-GM, which needs the two terms to be equal for equality.

I find that trying the AM-GM inequality when things are positive and you're stuck is helpful. This is a common problem in regional math olympiads.

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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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