Let $P(x)$ be a polynomial with only real roots and all coefficients equal to $\pm 1$. Prove that the degree of the polynomial is less than 4.

This is practice for Putnam, but I am not certain where to begin. I know I need to use inequalities. Could this be a roots of unity problem?

  • $\begingroup$ Well $x^4+1$ has no real roots, so there's a start. $\endgroup$ – Ian Coley Sep 18 '13 at 4:44
  • $\begingroup$ I think it's supposed to say "A polynomial has ONLY real roots". $\endgroup$ – Glen O Sep 18 '13 at 4:45
  • $\begingroup$ Yes allow me to correct that. The source for the problem also made that mistake. $\endgroup$ – Alex Sep 18 '13 at 4:49
  • $\begingroup$ There's also the fact that primitive $n$th roots of unity are not real after $n=2$ and they are roots of cyclotomic polynomials, which tend to have only coefficients $\pm1$. $\endgroup$ – Ian Coley Sep 18 '13 at 4:50
  • $\begingroup$ Just to state the obvious, all coefficients are supposed to be in $\{-1,0,1\}$. $\endgroup$ – Marc van Leeuwen Sep 18 '13 at 5:04

Let $P(x)=x^n+a_1x^{n-1}+...+a_n$ where all $a_i'$ s are $1$ or $-1$ and $x_1,x_2...x_n$ are all real roots of $P.$ By Viet's formulas $|x_1+x_2+...+x_n|=1$ and $x_1^2+x_2^2+...+x_n^2=(x_1+x_2+...x_n)^2-2\sum_{i<j}x_ix_j=1-2a_{2}=3.$ Now we can estimate $3=x_1^2+x_2^2+...x_n^2\ge n\sqrt[n]{x_1^2x_2^2...x_n^2}=n$ so, $n\le 3$ and the result follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.