What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root?


I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$ the equation becomes: $$X^{2}+aX+(b-2)=0$$ $$\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}$$ but I am not sure how to continue. If the polynomial has 1 root doesn't that is should be a double root? Are we counting multiplicity or not?

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    $\begingroup$ I think you may be thinking that if $x$ is a root, then so is $\frac 1 x$, so roots indeed come in pairs (with some special cases to consider). Once you have found $X$ you still have a quadratic to solve for $x$, so you need a condition on $X$ which makes $x$ real. $\endgroup$ – Mark Bennet Apr 27 '13 at 18:07
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    $\begingroup$ "At least one root" means it should have one, not excluding it has more (or one of them is double). $\endgroup$ – egreg Apr 27 '13 at 19:43
  • $\begingroup$ for any future readers, another solution can be found on this link. I unfortunately posted the same question two years later math.stackexchange.com/questions/1439501/… $\endgroup$ – Rishi Sep 18 '15 at 1:50

It seems to me that requesting the smallest possible value for $a^2+b^2$ is almost a red herring.


For $|a| \leq 4$ and $b > 2 |a| - 2$, the polynomial $$ p(x) = x^4 + ax^3 + bx^2 + ax + 1 $$ has no real zeros.


First set $b = 2|a| - 2 + \epsilon$, where $\epsilon > 0$. If we denote $\pm = \operatorname{sign} a$, so that $a = \pm |a|$, we have

$$ \begin{align} p(x) &= x^4 + ax^3 + (2|a|-2+\epsilon)x^2 + ax + 1 \\ &= x^4 - 2x^2 + 1 + x(ax^2 + 2|a|x + a) + \epsilon x^2 \\ &= (x^2-1)^2 + ax(x \pm 1)^2 + \epsilon x^2 \\ &= (x \pm 1)^2\left[ (x \mp 1)^2 + ax \right] + \epsilon x^2 \\ &= (x \pm 1)^2\left[ x^2 \pm (|a|-2)x + 1 \right] + \epsilon x^2 \\ &\geq (x \pm 1)^2 \min\left\{x^2 - 2x + 1,x^2+2x+1\right\} + \epsilon x^2 \\ &= (x \pm 1)^2\min\left\{(x-1)^2,(x+1)^2\right\} + \epsilon x^2 \\ &> 0. \end{align} $$


The points on the boundary of the region

$$ A = \{(a,b) \,\colon |a| \leq 4 \,\,\,\text{and}\,\,\,b > 2 |a| - 2\} $$

which have the smallest norm are

$$ (a,b) = \left(\pm \frac{4}{5},-\frac{2}{5}\right), $$

as shown in the following image.

enter image description here

It only remains to demonstrate that one of these points yields a polynomial with a real zero. In fact they both do:

$$ x^4 + \frac{4}{5} x^3 - \frac{2}{5} x^2 + \frac{4}{5} x + 1 $$

has a zero at $x=-1$ and

$$ x^4 - \frac{4}{5} x^3 - \frac{2}{5} x^2 - \frac{4}{5} x + 1 $$

has a zero at $x=1$.


The smallest value of $a^2 + b^2$ for which we can find some $a,b\in \mathbb R$ such that $p(x)$ has a real zero is $$ \left(\frac{4}{5}\right)^2 + \left(\frac{2}{5}\right)^2 = \frac{4}{5}. $$

Below is a plot of the $(a,b)$-plane which shows the region (in blue) where the polynomial $p(x)$ has at least one real root. Note that for $|a| \leq 4$ the boundary of this region is precisely $b = 2|a| - 2$ (shown in black), but for $|a| > 4$ it curves inward.

enter image description here

  • $\begingroup$ @AlexanderJones you're very welcome. It was a fun problem. $\endgroup$ – Antonio Vargas Apr 29 '13 at 15:39
  • $\begingroup$ By the way, I found this problem in a mathematical olympiad paper, I tried to do it for so long but couldn't, I don't know how maths olympiad people can do this so quickly :( $\endgroup$ – please delete me Apr 29 '13 at 15:55
  • $\begingroup$ @AlexanderJones, if it makes you feel any better I wouldn't have been able to solve it in an olympiad setting either. The first thing I did was use my computer to plot the second image in my post. Only once I could roughly see what was going on was I able to construct my answer. $\endgroup$ – Antonio Vargas Apr 29 '13 at 16:00

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