# Find all polynomials $f (x)$ such that $f (x^2+x+1)$ divides $f (x^3-1)$

I had come across a question in which involved finding polynomials (with real coefficients) satisfying the division criteria stated above. By inspection, it was easy to see that polynomials like $$x$$, $$x^2$$, $$x^3$$, etc. satisfied. So I went on to try a more general polynomial $$f (x)=ax^n$$ and it worked. I thought that if I'm able to prove that $$f (x)$$ can't have a non-zero root then it would suffice. Though I have found a solution that uses the 'assumption-contradiction' method (assuming that $$f (x)$$ has a non-zero root and showing a contradiction), but I was wondering that is there a technique that would actually allow us to solve the question (or questions of the same type) without guessing the answer first?

We will prove that all solutions are of a form $$ax^n$$, where $$n\in \mathbb{N}_0$$ and $$a\in \mathbb{R}$$.
Say exsist $$a_1\ne 0$$ such that $$f(a_1)=0$$. Then there exsist $$x_1\in \mathbb{C}$$ such that $$x_1^2+x_1+1=a_1$$ and $$|x_1-1|>1$$.
Such $$x_1$$ exsist since the equation $$x^2+x+1-a=0$$ has two solution $$x_1,x_2$$ for which $$x_1+x_2=-1$$. If $$|x_i-1|\leq 1$$ for each $$i$$, then we have by triangle inequality: $$2\geq |x_1-1|+|x_2-1|\geq |x_1+x_2-2| = 3$$
But then we have $$f(x_1^3-1) = k(x_1)f(a_1)=0$$ so $$a_2= x_1^3-1$$ is another root for $$f$$ and we can procede like this to get $$a_3, a_4,...$$. Since $$|a_2| = |x_1-1||a_1|>|a_1|$$ no two member of this sequence are equal.So we have infinite number of roots for $$f$$ which is a contradiction.
• Thank you very much. But as I mentioned earlier, does there exist a method which we can use for any general polynomial (or a more general polynomial)? I mean, here it was easy to guess $f(x)=ax^n$ but we may not always be able to do this. – Shashwat1337 Feb 8 '19 at 6:14