Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

one such point is (1,1,1)

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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)

Answer 9

got it??

Answer 10

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

Answer 11

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

Answer 12

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