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hsbhatt (6235)

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\text {Solve:} \\ \\    \frac {4x^2} {4x^2+1} = y \\ \\    \frac {4y^2}{4y^2+1} = z \\ \\    \frac {4z^2}{4z^2+1} = x

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sandy1990 (12)

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one could be x=y=z
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dinesh_ddt (163)

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x=y=z=0 is one soln

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hsbhatt (6235)

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I am looking for all solutions. x=y=z=0 is one. Are there any other. Your working should be able to exhaust all possibilities.

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sandy1990 (12)

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x=y=z=1/2

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sandy1990 (12)

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did u get it
i think it is the last possible solution

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hsbhatt (6235)

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did who get it?

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akhil_o (2709)

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i am not sure if we can do this, but by symmetry
x=y=z=a

so

4a2/(4a2+1)=a

so if a not equal to 0

4a=4a2+1

(2a-1)2=0
2a=1
or a=1/2

substituting a=0
works

so values of
a=x=y=z={0,1/2}

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sandy1990 (12)

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solved kya
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sandeepramesh (1247)

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WLOG assume 1>x>=y>=z>=0
Clearly from eqn1 and eqn3 making eqn3>=eqn1 we get z>=x which is a contradiction So x=y=z
 
so solving the quadratic eqns, x=y=z ={0,1/2}
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sandy1990 (12)

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thats wht i said

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sandeepramesh (1247)

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sorry, but where did you say that? im sorry i didnt see it :)
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sandy1990 (12)

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i think it was not posted on web
i told after the ans i gave 
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