Find all pairs $(x, y)$ such that $16^{x^{2}+y}+16^{x+y^{2}}=1 $

Question

Find all pairs $(x, y)$ of real numbers such that $$ 16^{x^{2}+y}+16^{x+y^{2}}=1 $$

I know this question has been answered multiple times but i have another solution which i have to validate.

So after taking $16^x16^y$ common from the equation we get

$$16^{x+y}=\dfrac{1}{16^{x(x-1)}+16^{y(y-1)}}$$

Now LHS is increasing while RHS is decreasing so equation can have at most one real root which can be find by putting x=y to get $(x, y)=(-1 / 2,-1 / 2)$ .

Anything wrong ?

Why is the RHS decreasing (in x)? In particular, what happens when $ x < \frac{1}{2}$, esp given that the solution involves $ x = - \frac{1}{2}$? — Calvin Lin, Oct 27 '20 at 05:22
What does it mean for a function in two variables to be increasing? — N. S., Oct 27 '20 at 05:22
What do you mean by an increasing (and decreasing) function of two variables? What is the partial ordering on $\mathbb{R}^2$? — Bumblebee, Oct 27 '20 at 05:24
As $16=2^4$ is a power of $2,$
one of the solutions will be $$x^2+y=x+y^2=-\dfrac14$$ — lab bhattacharjee, Oct 27 '20 at 05:27
@AlbusDumbledore OP is essentially asking for solution verification, not for solutions to the original problem. — Calvin Lin, Oct 27 '20 at 05:31
oops,i think i wrongly apply the fact that $a^x$ is increasing when $a>1$ ,thanks for pointing out. — Ishan, Oct 27 '20 at 05:32
@CalvinLin yes of course ,but now that the OP has got the mistake , this question becomes a duplicate — Albus Dumbledore, Oct 27 '20 at 05:34

Michael Rozenberg · Answer 1 · 2020-10-27T05:33:10.040

4

I think your reasoning is wrong.

Why the right side decreases?

How you define decreasing for the function of two variables?

By AM-GM $$1\geq2\sqrt{16^{x^2+y}\cdot16^{x+y^2}}=4^{x^2+y^2+x+y+\frac{1}{2}}=\sqrt2^{(2x+1)^2+(2y+1)^2}\geq1,$$ which gives $$x=y=-\frac{1}{2}.$$

edited Oct 27 '20 at 05:33

answered Oct 27 '20 at 05:26

Michael Rozenberg

194,933

Find all pairs $(x, y)$ such that $16^{x^{2}+y}+16^{x+y^{2}}=1 $

1 Answers1