Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

nuox95

  • 4 years ago

X+y = 2 2xy-(z^2)= find the interger solutions of the equation.

  • This Question is Closed
  1. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    is it 2xy-z^2=0??

  2. nuox95
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh wait , forgot to type it out, no, 2xy-z^2=1

  3. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if it is so ..then 2xy=z^2+1 (x+y)^2=4 x^2+y^2+2xy=4 x^2+y^2+z^2=3 so plot the sphere with radius sqrt(3) and find the points which has integer (x,y,z) triplet

  4. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    one such point is (1,1,1)

  5. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    only this is the solution..no other points are there.

  6. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u can prove it by this way x+y>=2sqrt(xy) sqrt(xy)<=1 as x+y=2 xy<=1 2xy<=2 z^2+1<=2 z^2<=1 z=1 only integer.. so 2xy=1^2+1=2 xy=1 s0 x=1 y=1 the solution.. (1,1,1) is the only possible solutin

  7. nuox95
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually I have the solution to the question which is x+y=2 and 2xy-z^2=1 leads to 2(x-1)^2+z^2=1, hence interger solutions are (1,1,1) and (1,1,-1) but i dont really get it, preparing for a math test which im not par with.

  8. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok....take my equation x^2+y^2+z^2=3 so it can be x^2=1, y^2=1, z^2=1 now x+y=2>0 so x,y>0 so x=y=1 but z^2=1 gives z=1,-1 so (1,1,1) and (1,1-1)

  9. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    got it??

  10. nuox95
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i get that z can be either -1 ,1 but how did you get x^2+y^2+z^2=3?

  11. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and also from second approach z^2<=1 z^2=1 z=1,-1 so 2xy=2 xy=1 and both x,y>0 so x=y=1

  12. dipankarstudy
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2xy=z^2+1 (x+y)^2=4 x^2+y^2+2xy=4 x^2+y^2+z^2=3

  13. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy