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Dec 8, 2014 at 10:29 comment added Michael Stoll 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.
Dec 8, 2014 at 2:30 history edited TZE CC BY-SA 3.0
simplified condition on P_l(x)
Dec 8, 2014 at 1:05 comment added TZE 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/…
Dec 8, 2014 at 0:26 vote accept TZE
S Dec 7, 2014 at 17:02 history suggested Michael Stoll
Edited tags.
Dec 7, 2014 at 16:22 review Suggested edits
S Dec 7, 2014 at 17:02
Dec 7, 2014 at 11:28 answer added Michael Stoll timeline score: 20
Dec 7, 2014 at 10:02 comment added Michael Stoll Did you observe any patterns (e.g. regarding the number of roots or their location) in your computation?
Dec 7, 2014 at 10:00 comment added Michael Stoll 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 $1$.
Dec 7, 2014 at 6:34 history asked TZE CC BY-SA 3.0