My Math Forum Equation...

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 June 6th, 2012, 06:12 PM #1 Member   Joined: May 2012 Posts: 86 Thanks: 0 Equation... Find all groups of three real numbers (x,y,z), such that (x-1)/y = (y-1)/z = (z-1)/x
 June 6th, 2012, 11:35 PM #2 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: Equation... $(x-1)/y =( y-1)/z = (z-1)/x = m , then x=my+1 , y=mz+1 , z=mx+1 upon substitution, (1-m^3)z = m^2+m+1 ==> z= 1/1-m this gives, x=y=z=1/(1-m) So solution is (1/(1-m),1/(1-m),1/(1-m)) m not = 1.$
June 7th, 2012, 02:38 PM   #3
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Re: Equation...

Quote:
 Originally Posted by karthikeyan.jp $(x-1)/y =( y-1)/z = (z-1)/x = m , then x=my+1 , y=mz+1 , z=mx+1 upon substitution, (1-m^3)z = m^2+m+1 ==> z= 1/1-m this gives, x=y=z=1/(1-m) So solution is (1/(1-m),1/(1-m),1/(1-m)) m not = 1.$
Your solution is simply x = y = z ? 0.

Are there solutions where they are not all equal.

June 7th, 2012, 02:48 PM   #4
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Re: Equation...

Quote:
Originally Posted by mathman
Quote:
 Originally Posted by karthikeyan.jp $(x-1)/y =( y-1)/z = (z-1)/x = m , then x=my+1 , y=mz+1 , z=mx+1 upon substitution, (1-m^3)z = m^2+m+1 ==> z= 1/1-m this gives, x=y=z=1/(1-m) So solution is (1/(1-m),1/(1-m),1/(1-m)) m not = 1.$
Your solution is simply x = y = z ? 0.

Are there solutions where they are not all equal.
Since it is symmetric on x,y,z, without loss of generality let x<=y<=z.
If we assume x<y<=z or x<=y<z or x<y<z; then $\frac{z-1}{x} > \frac{y-1}{z}$ which is impossible since both are equal to m.
The only remaining alternative is x=y=z, which obviously holds.

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