I have a question on my mind I am trying to solve. However I am stuck at a point. If you could help, I would be very pleased. $$\frac {x_2y_2z_2}{x_2y_2+y_2z_2+x_2z_2}=\frac {x_1y_1z_1}{x_1y_1+y_1z_1+x_1z_1}+6 $$ I have tried to solve this equation however I couldn't, by solving I mean finding $x_2,y_2,z_2$ in terms of $x_1,y_1,z_1$, and all of the terms must be integers. For instance for observation I tried doing it when $x_1=y_1=z_1$ and on solution became $x_2=x_1+18$ and similar for $y_1, z_1$. However I seemed to come to the conclusion that all three must be different: $x_1\neq x_2\neq x_3$, we can find solutions such as $x_2= y_1z_1+x_1+7$ however all of the terms $x_1,x_2, y_1,y_2, z_1,z_2$ must be positive integers. Thanks in advance!

  • $\begingroup$ Why the $+6$? Is there some reason you wouldn't be interested if there were a $+7$ on the right, instead? $\endgroup$ – Gerry Myerson Sep 1 '13 at 12:17
  • $\begingroup$ Like I said I am trying to solve another problem in which this equation pops up. However, if you do for 7 illustrating your method I can try to copy it for 6 too. $\endgroup$ – ciceksiz kakarot Sep 1 '13 at 12:25
  • $\begingroup$ "Like I said I am trying to solve another problem in which this equation pops up." So, what is the other problem? $\endgroup$ – Gerry Myerson Sep 1 '13 at 12:27
  • $\begingroup$ I am trying to construct three positive integers x,y,z that satisfies the equation $4xyz=(24m+1)(xy+yz+xz)$, and I want the solution for all m, so I thought induction would work, that's how I got to that equation. $\endgroup$ – ciceksiz kakarot Sep 1 '13 at 12:33
  • $\begingroup$ So, $${4\over24m+1}={1\over x}+{1\over y}+{1\over z}$$ You're working on the Erdos-Straus conjecture? $\endgroup$ – Gerry Myerson Sep 1 '13 at 12:39

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