Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$

Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $r\in(1,8)$. Any hints or solutions are welcome.

  • $\begingroup$ *For $ r=0 \ $? $\endgroup$ – For the love of maths Nov 11 '18 at 11:54
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    $\begingroup$ @Raptor rational number greater than 1 $\endgroup$ – Gareth Ma Nov 11 '18 at 12:09
  • $\begingroup$ Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z. $\endgroup$ – William Elliot Nov 11 '18 at 22:16

x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.

Often there will be no solution. r = 7 for example.

Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.

Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.

Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.

Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.


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