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Mathematics 38 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

one such point is (1,1,1)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

got it??

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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