Prove that there isn't a polynomial with $\text {f(x)}^{13} = {(x-1)}^{143}+(x+1)^{2002}$

We can easily find out that $\text {deg}(f) = 154$


  • $\begingroup$ I think you didn't want a $13$ in your suggestion. $\endgroup$ May 14, 2013 at 15:05
  • 1
    $\begingroup$ What field are the coefficients of $f$ in? $\endgroup$ May 14, 2013 at 15:21
  • $\begingroup$ Yes @sharkos, thanks. Assume they are in a Field ( f en R[x]) $\endgroup$
    – Mike B
    May 14, 2013 at 15:24

2 Answers 2


Hint: Consider the coefficients of $x^0,x^1$ on both sides of the equation. This suffices.

  • $\begingroup$ nice hint${}{}{}{}+1$ $\endgroup$
    – Amr
    May 14, 2013 at 15:23
  • $\begingroup$ I am very stuck trying to wrap my mind around this. Obviously f(0) = 0, so x^0 coefficient is 0. But for x^1 do you expand on binomials? The only path that makes sense to me is starting with f(x)=xf'(x), and mcd( x-1,x+1,x) = 1 $\endgroup$
    – Mike B
    May 14, 2013 at 15:43
  • $\begingroup$ You could expand the first binomial terms, yes. If you prefer to think in terms of derivatives, you should consider the derivative of the whole expression at 0. This will give a contradiction. $\endgroup$ May 14, 2013 at 15:55
  • $\begingroup$ Thank you! This makes sense now $\endgroup$
    – Mike B
    May 14, 2013 at 16:03
  • $\begingroup$ Why the downvote, whoever did it? If there's a problem it's only decent to let the OP know at least! $\endgroup$ May 15, 2013 at 1:06

This, I think, is a nice problem, and deserves a hint rather than an answer.

See what happens with some values of $x$ which are easy to calculate. What does this tell you about the form of $f(x)$?

  • $\begingroup$ I cannot figure out values easy to calculate. Except for the obvious f(0) we can only know that for example f(-1) is "very negative" or f(1) "very positive" $\endgroup$
    – Mike B
    May 14, 2013 at 15:38
  • $\begingroup$ You have an explicit value for each of these. Pick the easiest one - what do you know about a polynomial where $f(0)=0$ - I see from comments to Sharkos that you get that. $f(x)=xg(x)$ where $g$ is a polynomial. Now raise that to the power 13. And consider what Sharkos suggested. $\endgroup$ May 14, 2013 at 15:57
  • $\begingroup$ Thank you, I understand now since sharko's post has accumulated more +1s I will also mark it as preffered answer, although yours is equally good hint! $\endgroup$
    – Mike B
    May 14, 2013 at 16:02

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