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Dec 5, 2017 at 8:26 comment added Peter Mueller @GerryMyerson You are right, I didn't know about this problem.
Dec 4, 2017 at 22:30 comment added Gerry Myerson Isn't $r=d$ just the Tarry-Escott problem?
Dec 4, 2017 at 20:42 comment added Johann Franke Wow, this is pretty awesome! Thank you!
Dec 4, 2017 at 20:41 vote accept Johann Franke
Dec 4, 2017 at 16:57 comment added Peter Mueller I'm wondering if actually $r=d$ is possible (not by this beautiful pigeon hole argument of course). At least it works for $d\le6$. For instance, for $d=6$, set $P(X)=(X-1)(X-2)(X-10)(X-12)(X-20)(X-21)$ and $Q(X)=X(X-5)(X-6)(X-16)(X-17))(X-22)$. Then $P(X)-Q(X)=100800$.
Dec 4, 2017 at 14:20 history answered js21 CC BY-SA 3.0