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  • $\begingroup$ The condition on $P_l$ seems to be unnecessarily complicated. It is equivalent to $P_l$ not having any roots in ${\mathbb Z}/l{\mathbb Z}$ except possibly $0$ and/or $1$. $\endgroup$ Commented Dec 7, 2014 at 10:00
  • $\begingroup$ Did you observe any patterns (e.g. regarding the number of roots or their location) in your computation? $\endgroup$ Commented Dec 7, 2014 at 10:02
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    $\begingroup$ Wow! Thank you for the complete answer! Yes, I did. There were always at least two solutions excluding $0, 1.$ and they appeared in pairs $x, x^{-1}.$ Here is the Sage computation with the output: drive.google.com/file/d/0B_kUj8Mvy6_NUlNaLVg4VHJ0V1U/… $\endgroup$ Commented Dec 8, 2014 at 1:05
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    $\begingroup$ You are welcome. -- I was suspecting that cube roots of unity might play a role (from the condition $l \equiv 1 \bmod 3$); looking at the first few cases confirmed that. $\endgroup$ Commented Dec 8, 2014 at 10:29