Question

here's a neat olympiad problem

Answer 1

\[16^{x^2+y}+16^{x+y^2}=1\]Find all pairs \((x,y)\) satisfying this.

Answer 2

a very childlike examination suggests that because the expression is symmetric we can try solving for \(x=y\) so \(2^{4x^2 + 4x}=2^{-1}\) and thus \(x=y=-1/2\).

the real challenge is finding other solutions.

Answer 3

i know how to do it anyways because i did it already before
x^2+y+x+y^2+1/2=(x+1/2)^2 + (y+1/2)^2 >=0
x^2+y+x+y^2 >=-1/2
1=16^(x^2+y) + 16^(x+y^2) >=2*(16^(x^2+y) * 16^(x+y^2))^(1/2)=2*(16^(x^2+x+y+y^2)^(1/2) >=2*16^(-1/4)=1

(x+1/2)^2 + (y+1/2)^2=0

thus only (-1/2, -1/2) is solution

Answer 4

it is known Olympic problem

Answer 5

these guys have same solution

http://olympiads.hbcse.tifr.res.in/uploads/rmo-2011-solutions

Answer 6

Boo-hoo.

Answer 7

That's where I got the problem. One should think about a problem before resorting to search engines.

Answer 8

lol i know it

i had same problem on my olympic,

Answer 9

just it was long ago, wanted to check if i'm correct