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siddhantsharanBest ResponseYou've already chosen the best response.0
\[16^{x^2 + y} + 16^{y^2 + x} = 1\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
are those real numbers?
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
2^4(x^2+y) + 2^4(y^2+x) = 2^1 + 2^1
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
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
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
That is ONE SOLUTION. There may be more. Yes, they're real.
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
@mukushla im clueless plz give hint or something :)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
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
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Howd you make that conclusion?
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
I know I;m being pretty stupid somewhere. ;/ Still, I dont see it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1349455630674:dw
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Both x and y have to be negative.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
lol ... i meant that.sorry i forgot to put  on the left. dw:1349455681739:dw
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Why does a symmetric situation imply that? Can you prove that THAT will be the only solution?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
if you change x and y in that equation, there isn't any change ... so x and y must be interchangeable.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1349455922847:dw
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
I agree that that case seems highly plausable, but how do we prove it?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Let us define a function on R^2 dw:1349456340587:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago
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