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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
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