Equation with x,y as exponential functions
1. The problem statement, all variables and given/known data
This question was asked in the Indian National Maths Olympiad. The question is: Find all the possible real ordered pairs of (x,y) for equation 16^[(x^2) + y] + 16^[x + (y^2)]=1 2. Relevant equations That was the only equation. 3. The attempt at a solution Using 0.5 + 0.5 =1, (x^2) + y = 0.25 and (y^2) + x= 0.25 (Because [16^(0.25)]=0.5 , I'm using decimals here instead of fractions) Solving them gives one real ordered pair of (0.5,0.5). But how do I know if that's the only answer? 
Re: Equation with x,y as exponential functions
This is my first thread ever, by the way. Tell me if I did something wrong.

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Re: Equation with x,y as exponential functions
Was up till 3 a.m. here last night for a family arrival, and wrote last post just before going to bed. Had thought to write 'is it plausible that the answer is one or a few pairs of numbers like that?'.
Woke about 7, and in a bit saw ± what you are supposed to do. It starts looking nice but at the end the algebra got looking ugly. Then realised you are supposed to describe answer geometrically, not algebraically. :wink: Have not needed to write anything down. Wondering whether to go back to bed or have breakfast. 
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Re: Equation with x,y as exponential functions
I tried over two days to see what wolframapha would make of this, but it's not functioning. Anyone know whether the free version has been decommissioned? :grumpy:

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(It might be possible to plot a graph of 16^((x^2) + y) + 16^((y^2) + x) = z, and then intersecting the surface with z=1. But I couldn't find anything that would plot something like that. They all accept y as a function of x. This doesn't really get us anywhere. ) Is there anyway to find the minimum or maximum value of 16^((x^2) + y) + 16^((y^2) + x)? That might help. Possible ways to do so might be using the Arithmetic Mean >= Geometric mean >= Harmonic mean property. Or maybe by just differentiating it and putting dy/dx=0. But I couldn't find anything there. Do tell me if anyone finds anything. 
Re: Equation with x,y as exponential functions
I'm having quite a lot of trouble with this problem, it's really got me stumped. But then again, all the Olympiad questions seem to give me a hard time :redface:
Since we have an exponential of the form [itex]a^b+a^c=1[/itex] and we know that [itex]a^b>0, a^c>0[/itex] then what we must have is that [itex]b<0, c<0[/itex] So [tex]x^2+y<0[/tex] [tex]x+y^2<0[/tex] And this gives us the inequality [tex]1<x<0[/tex] [tex]\sqrt{x}<y<x^2[/tex] But from here I can't think of a way to further narrow down our possible solution set. 
Re: Equation with x,y as exponential functions
Hey, Mentallic. Nice try...
This did help a bit... http://www.wolframalpha.com/input/?i...B+1%3C+y+%3C0 
Re: Equation with x,y as exponential functions
Actually, However, that doesn't narrow down to it.

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Given this, and in light of my inability to get it to draw even a straightline graph, I think the owners must have made it regionally selective, so it's ignoring certain IP ranges. When I go to http://www.wolframalpha.com I am presented with a totally blank page. Anyone else? Maybe this facility comes under the umbrella of "National security ..........?" http://i.imgur.com/L5zX9.gif 
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Haha! Probably. 
Re: Equation with x,y as exponential functions
Hmmmm, no solutions then?
(0.5,0.5) only? 
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You might therefore try rotating the thing counter clockwise by 45° , then it is symmetrical around the new X axis f(X) = f(X) and see if that inspires anything. 
Re: Equation with x,y as exponential functions
Well, if it's a curve, why wouldn't anyone plot it?
Wolfram Alpha plots a curve for 16^[(x^2) + y] + 16^[x + (y^2)]= anything more than 1. Maybe whatever the curve is, it gets reduced to a point when 16^[(x^2) + y] + 16^[x + (y^2)]= 1 
Re: Equation with x,y as exponential functions
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