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Jul 24, 2015 at 22:27 vote accept Adam Sheffer
Jul 24, 2015 at 3:16 comment added Will Sawin @AdamSheffer We can do $a(x) = b(y)$ where $b'(y)$ has no real and $a(x)=z$ has $d$ roots for $z$ in some interval. Then perturbing $b$ by some polynomial with $d$ roots in that interval gives $d^2$, for $d$ odd, or $d(d-1)$, for $d$ even.
Jul 23, 2015 at 23:45 comment added Adam Sheffer Very nice! Thank you. In general this symmetry trick gives $2d$ intersection points, which is still far from $d^2$. Do you think that it is possible to have more intersection points than that?
Jul 23, 2015 at 23:40 vote accept Adam Sheffer
Jul 23, 2015 at 23:42
Jul 23, 2015 at 23:10 history answered Will Sawin CC BY-SA 3.0