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Mar 29, 2018 at 14:07 vote accept Alexander Kalmynin
Mar 29, 2018 at 12:54 vote accept Alexander Kalmynin
Mar 29, 2018 at 12:56
Mar 29, 2018 at 12:51 vote accept Alexander Kalmynin
Mar 29, 2018 at 12:54
Mar 29, 2018 at 12:34 answer added Joe Silverman timeline score: 10
Mar 27, 2018 at 20:10 comment added Joe Silverman @SylvainJULIEN In fact, it seems that $f_n(x)\mid f_{n+1}(x)$. What seems to be happening is that at each stage, the numerator is picking up an additional factor that is equal to the previous denominator. So I would start by asking if $q_n(x)$, which I'll define to be the denominator of $p_n(x)$, has square discriminant. The first few $q_n(x)$ are irreducible in $\mathbb Q[x]$, so it's interesting that their discriminants are squares.
Mar 27, 2018 at 20:06 comment added Sylvain JULIEN It seems judging from your list that for all $ n $ one has $ D_{n}\mid D_{n+1} $ .
Mar 27, 2018 at 19:28 history edited Joe Silverman CC BY-SA 3.0
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Mar 27, 2018 at 19:14 history asked Alexander Kalmynin CC BY-SA 3.0