A community for students.
Here's the question you clicked on:
 0 viewing
nuox95
 3 years ago
X+y = 2
2xy(z^2)=
find the interger solutions of the equation.
nuox95
 3 years ago
X+y = 2 2xy(z^2)= find the interger solutions of the equation.

This Question is Closed

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1is it 2xyz^2=0??

nuox95
 3 years ago
Best ResponseYou've already chosen the best response.0oh wait , forgot to type it out, no, 2xyz^2=1

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1one such point is (1,1,1)

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1only this is the solution..no other points are there.

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1u 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

nuox95
 3 years ago
Best ResponseYou've already chosen the best response.0actually I have the solution to the question which is x+y=2 and 2xyz^2=1 leads to 2(x1)^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.

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1ok....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 (1,11)

nuox95
 3 years ago
Best ResponseYou've already chosen the best response.0i get that z can be either 1 ,1 but how did you get x^2+y^2+z^2=3?

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.1and 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

dipankarstudy
 3 years ago
Best ResponseYou've already chosen the best response.12xy=z^2+1 (x+y)^2=4 x^2+y^2+2xy=4 x^2+y^2+z^2=3
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.