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Solve for (x,y) Such that

Mathematics
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\[16^{x^2 + y} + 16^{y^2 + x} = 1\]
are those real numbers?
2^4(x^2+y) + 2^4(y^2+x) = 2^-1 + 2^-1

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Other answers:

i tempted to start like that... and maybe equate exponents, and getting (x,y) = (-1/2, -1/2) . plz see if any flaws in my logic... guys
That is ONE SOLUTION. There may be more. Yes, they're real.
@mukushla im clueless plz give hint or something :)
clearly x<0 and y<0 and the equation is symmetric so suppose\[y\ge x \]conclude that just y=x gives us an answer
Howd you make that conclusion?
I know I;m being pretty stupid somewhere. ;/ Still, I dont see it.
|dw:1349455630674:dw|
Both x and y have to be negative.
lol ... i meant that.sorry i forgot to put - on the left. |dw:1349455681739:dw|
Why does a symmetric situation imply that? Can you prove that THAT will be the only solution?
if you change x and y in that equation, there isn't any change ... so x and y must be interchangeable.
|dw:1349455922847:dw|
I agree that that case seems highly plausable, but how do we prove it?
that gives one solution ... since x and y are interchangeable at x=y we just have to show that no other solution exist or if it exists then find them.
Let us define a function on R^2 |dw:1349456340587:dw|
Woops!! I made an error, ... earlier ... the f(x, y) goes to heaven for x, y -> -inf ... there is only one solution ... the minimum value of this surface is at x=-0.5 and y=-0.5 http://www.wolframalpha.com/input/?i=Minimize+16^%28x^2%2By%29+%2B+16^%28x%2By^2%29+-1

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