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# Understand the problem

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

Algebra
Easy
##### Suggested Book
Problem Solving Strategies by Arthur Engel

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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