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Jun 18, 2019 at 19:52 answer added Will Sawin timeline score: 11
Jun 14, 2019 at 19:16 comment added R.P. @KonstantinosGaitanas That is clear. My point was that the formula given earlier in the post does not apply in that case.
Jun 14, 2019 at 19:15 answer added R.P. timeline score: 20
Jun 14, 2019 at 19:11 comment added Konstantinos Gaitanas @RP_ so, $M=-1$.
Jun 14, 2019 at 19:02 comment added R.P. Very probably something very much like what you claim is true, except it can't be completely correct since $P(x)=x^2+x$ also works, which satisfies $P(x) = P(-1-x)$. There is a subtlety in that the symmetry-axis can oocur for a half-integral $x$-value.
Jun 14, 2019 at 18:53 comment added Seva It seems easy to see at least that $P$ must be of even degree.
S Jun 14, 2019 at 17:38 history suggested ssx
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Jun 14, 2019 at 17:34 review Suggested edits
S Jun 14, 2019 at 17:38
Jun 14, 2019 at 17:15 history edited apple CC BY-SA 4.0
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Jun 14, 2019 at 16:26 history asked apple CC BY-SA 4.0