## nuox95 4 years ago X+y = 2 2xy-(z^2)= find the interger solutions of the equation.

1. dipankarstudy

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

2. nuox95

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

3. dipankarstudy

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

one such point is (1,1,1)

5. dipankarstudy

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

6. dipankarstudy

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

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

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

got it??

10. nuox95

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

11. dipankarstudy

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

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