Understand the problem

Find all polynomials with real coefficients $P(x)$ such that $P(x^2+x+1)$ divides $P(x^3-1)$.

Source of the problem
Indian National Mathematical Olympiad 2018
Difficulty Level
Suggested Book
Problem Solving Strategies by Arthur Engel

Start with hints

Do you really need a hint? Try it first!

Show that, given a zero of P, it is possible to find another zero.
Whenever such a property holds (as in hint 1), it is a good idea to use the extremal principle (that is, looking at extremal elements).
Among all (possibly complex) zeros of P, consider the one with the maximum absolute value. Show that, for most choices of P, it is possible to construct another zero with larger absolute value. As that is absurd, a large number of options for P are eliminated.

As P(x^3-1) is divisible by P(x^2+x+1), there exists a polynomial Q(x) such that P(x^2+x+1)Q(x)=P(x^3-1). Let z be a zero of P. If x^2+x+1=z then x=\frac{-1\pm\sqrt{1+4(z-1)}}{2}, i.e. x^3-1= (x-1)(x^2+x+1)=z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right). The given equation means that z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right) are also zeroes of P. Claim One of the two numbers \left|\frac{-3\pm\sqrt{4z-3}}{2}\right| is greater than 1.   Proof Suppose that they are both \le 1. Then we have   2\ge |-3+\sqrt{4z-3}|   2\ge |-3-\sqrt{4z-3}| Adding these two inequalities we get 4\ge |-3+\sqrt{4z-3}|+|-3-\sqrt{4z-3}|\ge |(-3+\sqrt{4z-3})+(-3-\sqrt{4z-3})|=6 which is absurd. Hence at least one of them has to be greater than 1.     The claim means that, if z is non-zero, then at least one of the two zeroes z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right) has absolute value greater than |z|.

Now let z=z_0, the zero of P with the largest absolute value. The procedure mentioned above can be used to construct a zero with absolute value bigger than |z_0|, which is absurd unless z_0=0. As z_0 is largest among the zeroes in absolute value, this means that P does not have any nontrivial zeroes. Hence P(x)=ax^n for some a\in\mathbb{R} and n\in\mathbb{N}.

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