Equation with x,y as exponential functions

  1. 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?
  2. jcsd
  3. This is my first thread ever, by the way. Tell me if I did something wrong.
  4. epenguin

    epenguin 2,126
    Homework Helper
    Gold Member

    Have you checked that that is a solution?
  5. epenguin

    epenguin 2,126
    Homework Helper
    Gold Member

    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.
    Last edited: Mar 10, 2012
  6. (-0.5,-0.5) fits in perfectly. But the question is to find all possible real ordered pairs. How do I do that?
  7. What do you mean???
  8. NascentOxygen

    Staff: Mentor

    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:
  9. I've tried that already. It would just do a differentiation. The problem can only be solved by a human.

    (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.
  10. Mentallic

    Mentallic 3,661
    Homework Helper

    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]


    And this gives us the inequality
    But from here I can't think of a way to further narrow down our possible solution set.
  11. Actually, However, that doesn't narrow down to it.
  12. epenguin

    epenguin 2,126
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    Gold Member

    I mean - that I had misread the question :biggrin: (missed the ^) :blushing:
  13. NascentOxygen

    Staff: Mentor

    You get some sort of result when you click on this link, do you? As I said previously, wolframalpha seems non-functioning to me. I clicked on your link and for 10 mins watched it spinning endlessly on "Computing ..."

    Given this, and in light of my inability to get it to draw even a straight-line 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 ..........?" [​IMG]

  14. Haha! Probably.
  15. Hmmmm, no solutions then?

    (-0.5,-0.5) only?
  16. epenguin

    epenguin 2,126
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    Gold Member

    Can hardly be. One equation relating two variables is a continuous curve in general. The present curve must be symmetrical about the 'diagonal' line x = y and go through that point. The curve might have more than one branch but if so no other branch cuts that line. Don't know what else they want you to say about it.

    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.
    Last edited: Mar 15, 2012
  17. 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
  18. epenguin

    epenguin 2,126
    Homework Helper
    Gold Member

    Can you show us what the family looks like, so we see what happens as you reduce this something to 1?
  19. I can't.
  20. Mentallic

    Mentallic 3,661
    Homework Helper

    I've just been playing with things for a bit, and just thought I'd dumb things down for simplicity, which could also mean I dumbed things down a little too far.
    If we look at the exponents only, and take the addition of them we have a function

    [tex]=\left( x+\frac{1}{2} \right)^2+\left( y+\frac{1}{2} \right)^2-\frac{1}{2}[/tex]

    So we clearly have a symmetry about y=x (as we already knew) and the minimum value f can take is -1/2.

    At this point [itex](x,y)=(-1/2,-1/2)[/itex] which is where the minimum of f occurs, we found that


    And so if we take some other value, say, [itex]x=\frac{-1}{2}+k[/itex], [itex]y=\frac{-1}{2}-k[/itex], [itex]k\neq0[/itex] ***
    then with these values we have



    Therefore, g becomes



    And for all k


    So we only have the minimum at (-1/2,-1/2) which means this is the only combination (x,y) that satisfies [itex]g(x,y)=1[/itex]

    *** I'm uneasy about making this assumption that if we take some small linear increase of k in one variable, we should decrease the other variable with the same magnitude k.
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