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How to find all pairs $(x,z)$ of integers for which $2(z+1)^3$ is divisible by $xz-1$

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Have you tried compiling numerous examples and searching them for patterns? –  Gerry Myerson Oct 26 '12 at 6:32

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Here's all the pairs of $(x,z)$ I could find for positive $x,z$. On the right is $2(z+1)^3/(xz-1)$

\begin{array}{cc} \left(1,2\right) & 54 \\ \left(1,3\right) & 64 \\ \left(1,5\right) & 108 \\ \left(1,9\right) & 250 \\ \left(1,17\right) & 729 \\ \left(2,1\right) & 16 \\ \left(2,2\right) & 18 \\ \left(2,5\right) & 48 \\ \left(2,14\right) & 250 \\ \left(3,1\right) & 8 \\ \left(3,3\right) & 16 \\ \left(3,11\right) & 108 \\ \left(3,43\right) & 1331 \\ \left(5,1\right) & 4 \\ \left(5,2\right) & 6 \\ \left(5,5\right) & 18 \\ \left(5,11\right) & 64 \\ \left(5,29\right) & 375 \\ \left(9,1\right) & 2 \\ \left(9,9\right) & 25 \\ \left(9,14\right) & 54 \\ \left(11,3\right) & 4 \\ \left(11,5\right) & 8 \\ \left(11,35\right) & 243 \\ \left(14,2\right) & 2 \\ \left(14,9\right) & 16 \\ \left(17,1\right) & 1 \\ \left(29,5\right) & 3 \\ \left(29,69\right) & 343 \\ \left(35,11\right) & 9 \\ \left(43,3\right) & 1 \\ \left(69,29\right) & 27 \end{array}

See anything?

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I couldn't see the pattern . –  abt Oct 26 '12 at 7:26

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