Suppose p is the polynomial p(z)=az4+bz3+cz2+dz+e, where a, b, c, d, e are complex numbers with a=0. Prove that there is a complex number h such that p(h+z)=p(h−z) for all complex numbers z, if and only if b3−4abc+8a2d=0.
Solution
We have 0=p(h+z)−p(h−z)=a((h+z)4−(h−z)4)+b((h+z)3−(h−z)3)+c((h+z)2−(h−z)2)+d((h+z)−(h−z))=8a(h3z+hz3)+2b(3h2z+z3)+4chz+2dz=z(8ah3+6bh2+4ch+2d)+z3(8ah+2b). This is satisfied for all z∈C iff 8ah3+6bh2+4ch+2d=0and8ah+2b=0. The second equation implies h=−4ab. Substituting this into the first of these equations, we obtain, after multiplying by 4a2, b3−4abc+8a2d=0. Thus, this condition is necessary. Conversely, if it holds, then, with h=−b/(4a) we see that 8ah3+6bh2+4ch+2d=0 and 8ah+2b=0 hold.
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