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

nuox95Best ResponseYou've already chosen the best response.0
oh wait , forgot to type it out, no, 2xyz^2=1
 2 years ago

dipankarstudyBest ResponseYou've already chosen the best response.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
 2 years ago

dipankarstudyBest ResponseYou've already chosen the best response.1
one such point is (1,1,1)
 2 years ago

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

dipankarstudyBest ResponseYou've already chosen the best response.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
 2 years ago

nuox95Best ResponseYou've already chosen the best response.0
actually 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.
 2 years ago

dipankarstudyBest ResponseYou've already chosen the best response.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 (1,11)
 2 years ago

nuox95Best ResponseYou've already chosen the best response.0
i get that z can be either 1 ,1 but how did you get x^2+y^2+z^2=3?
 2 years ago

dipankarstudyBest ResponseYou've already chosen the best response.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
 2 years ago

dipankarstudyBest ResponseYou've already chosen the best response.1
2xy=z^2+1 (x+y)^2=4 x^2+y^2+2xy=4 x^2+y^2+z^2=3
 2 years ago
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