So I was going through my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $\pm 1$ as all its coefficients.

I tried proving it using calculus, by showing that at least one consecutive maxima and minima will lie either above or below the x axis, but couldn't solve it using that.

I also tried using Descartes Rule of Signs but couldn't solve it with that too. Any help?

  • 3
    $\begingroup$ Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two. $\endgroup$ – Arthur Jul 16 '18 at 15:25
  • 1
    $\begingroup$ I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially. $\endgroup$ – hardmath Jul 16 '18 at 15:27
  • 1
    $\begingroup$ To be precise, there are $32$ such polynomials $\endgroup$ – asdf Jul 16 '18 at 15:27
  • 4
    $\begingroup$ @asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check. $\endgroup$ – Arthur Jul 16 '18 at 15:28
  • $\begingroup$ True, I thought you aren't narrowing the cases in any way before bashing it out $\endgroup$ – asdf Jul 16 '18 at 15:29

Let $f(x)$ be any quartic polynomial with coefficients from $\{ -1, +1 \}$. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.

$$f(x) = x^4 + ax^3 + bx^2 + cx + d\quad\text{ with }\quad a,b,c,d \in \{ -1, +1 \}$$

If $f(x)$ has $4$ real roots $\lambda_1,\lambda_2,\lambda_3,\lambda_4$, then by Vieta's formula, we have

$$\sum_{i=1}^4 \lambda_i = -a, \sum_{1\le i < j\le 4} \lambda_i\lambda_j = b \quad\text{ and }\quad\prod_{i=1}^4 \lambda_i = d$$ Notice $$\sum_{i=1}^4 \lambda_i^2 = \left(\sum_{i=1}^4\lambda_i\right)^2 - 2\sum_{1\le i < j \le 4}\lambda_i\lambda_j = a^2 - 2b = 1 -2b$$

Since $\sum_{i=1}^4 \lambda_i^2 \ge 0$, we need $b = -1$. As a result, $$\sum_{i=1}^4 \lambda_i^2 = 3$$ By AM $\ge$ GM, this leads to

$$\frac34 = \frac14\sum_{i=1}^4 \lambda_i^2 \ge \left(\prod_{i=1}^4 \lambda_i^2\right)^{1/4} = (d^2)^{1/4} = 1$$ This is impossible and hence $f(x)$ cannot has 4 real roots.

  • $\begingroup$ You are not using at any point that the sum of product of triples is $\pm1$, so the claim can be generalised? $\endgroup$ – asdf Jul 16 '18 at 15:52
  • $\begingroup$ @asdf yes, the proof doesn't need the value of $c$. $\endgroup$ – achille hui Jul 16 '18 at 15:53
  • $\begingroup$ Nice, $(+1)$ from me $\endgroup$ – asdf Jul 16 '18 at 15:53
  • $\begingroup$ Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate. $\endgroup$ – James Adams Jul 16 '18 at 15:56
  • 1
    $\begingroup$ @JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details. $\endgroup$ – achille hui Jul 17 '18 at 16:11

It can be assumed WLOG that the leading coefficient is $\,+1\,$, so $\,P(x)=x^4\pm x^3\pm x^2\pm x\pm 1\,$.

  • Then $\,P''(x)=12x^2 \pm 6x \pm 2\,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $\,P(x)=x^4\pm x^3 - x^2\pm x\pm 1\,$.

  • $P(x)\,$ has all the roots real iff $\,x^4 P\left(\frac{1}{x}\right)\,$ has all real roots. By the same argument as above, the constant term of $\,P(x)\,$ must have opposite sign as the coefficient of $\,x^2\,$.

This leaves $4$ cases to check $\,P(x)=x^4\pm x^3 - x^2\pm x+1\,$.

[ EDIT ]

  • $P(x)\,$ has all the roots real iff $\,P\left(-x\right)\,$ has all real roots, so it is enough to consider the case where the coefficient of $\,x^3\,$ is $+1$.

This leaves $2$ cases to check $\,P(x)=x^4+ x^3 - x^2\pm x+1\,$.

  • $\begingroup$ @RossMillikan On the other hand, it is enough to consider one of the cases $\,\pm x^3\,$, as noted in the latest edit. $\endgroup$ – dxiv Jul 16 '18 at 16:08

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.