Maths Olympiad Prep

Library / /155 of 462

Algebra Difficulty 5.4 AIME, harder Prove it Ireland

Suppose pp is the polynomial
p(z)=az4+bz3+cz2+dz+e, p(z) = az^4 + bz^3 + cz^2 + dz + e,
where aa, bb, cc, dd, ee are complex numbers with a0a \neq 0. Prove that there is a complex number hh such that p(h+z)=p(hz)p(h+z) = p(h-z) for all complex numbers zz, if and only if
b34abc+8a2d=0. b^3 - 4abc + 8a^2d = 0.

Solution

We have
0=p(h+z)p(hz)=a((h+z)4(hz)4)+b((h+z)3(hz)3)+c((h+z)2(hz)2)+d((h+z)(hz))=8a(h3z+hz3)+2b(3h2z+z3)+4chz+2dz=z(8ah3+6bh2+4ch+2d)+z3(8ah+2b). \begin{aligned} 0 &= p(h+z) - p(h-z) \\ &= a((h+z)^4 - (h-z)^4) + b((h+z)^3 - (h-z)^3) \\ &\quad + c((h+z)^2 - (h-z)^2) + d((h+z) - (h-z)) \\ &= 8a(h^3z + hz^3) + 2b(3h^2z + z^3) + 4chz + 2dz \\ &= z(8ah^3 + 6bh^2 + 4ch + 2d) + z^3(8ah + 2b). \end{aligned}
This is satisfied for all zCz \in \mathbb{C} iff
8ah3+6bh2+4ch+2d=0and8ah+2b=0. 8ah^3 + 6bh^2 + 4ch + 2d = 0 \quad \text{and} \quad 8ah + 2b = 0.
The second equation implies h=b4ah = -\frac{b}{4a}. Substituting this into the first of these equations, we obtain, after multiplying by 4a24a^2,
b34abc+8a2d=0. b^3 - 4abc + 8a^2d = 0.
Thus, this condition is necessary. Conversely, if it holds, then, with h=b/(4a)h = -b/(4a) we see that 8ah3+6bh2+4ch+2d=08ah^3 + 6bh^2 + 4ch + 2d = 0 and 8ah+2b=08ah + 2b = 0 hold.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.