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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Shashwat1337 Feb 8 at 6:14

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