Kvant Math Problem 469

Let $P(x)=x^4+ax^3+bx+c$ have four distinct real roots $r_1<r_2<r_3<r_4$.

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Problem

  1. The equation $x^4+ax^3+bx+c=0$ has four distinct real roots. Prove that $ab\lt0$.
  2. The equation $x^n+a_1x^{n-1}+\ldots+a_{k-1}x^{n-k+1}+a_{k+1}x^{n-k-1}+\ldots+a_n=0$ has $n$ distinct real roots. Prove that $a_{k-1}a_{k+1}\lt0$.

V. V. Vavilov

Exploration

Let $P(x)=x^4+ax^3+bx+c$ have four distinct real roots $r_1<r_2<r_3<r_4$. By Viète,

$$r_1+r_2+r_3+r_4=-a,\qquad r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4=-b.$$

Thus $ab<0$ is equivalent to showing that

$$(r_1+r_2+r_3+r_4),(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)<0.$$

A direct sign analysis of these symmetric sums is unstable because root signs may vary. A more structural route is to exploit interlacing roots of $P$ and $P'$, then construct a derived polynomial whose real-rootedness forces a sign condition on coefficients.

For the quartic case, the cubic

$$Q(x)=xP'(x)-4P(x)$$

eliminates the $x^4$ term and produces a cubic with coefficients depending only on $a,b,c$. If $Q$ has three real roots, then its depressed cubic form forces a sign constraint equivalent to $ab<0$.

The key difficulty is proving that $Q$ indeed has three real roots. This must be done using sign changes at critical points of $P$, not by assuming Sturm theory.

Problem Understanding

This is a Type A classification problem.

For the quartic, we must prove that if $x^4+ax^3+bx+c$ has four distinct real roots, then $ab<0$.

For the general polynomial with one missing coefficient $a_k=0$, we must prove that the neighboring coefficients satisfy $a_{k-1}a_{k+1}<0$.

The core difficulty is to convert information about real-rootedness into a rigid sign constraint on specific symmetric combinations of roots, without appealing to unproven monotonicity or coefficient heuristics.

Proof Architecture

Let $P$ be the given polynomial.

We construct the auxiliary polynomial $Q(x)=xP'(x)-nP(x)$ in degree $n$, which reduces to degree $n-1$.

We prove that $Q$ has $n-1$ distinct real roots by combining Rolle’s theorem for $P$, sign alternation of $P$ at critical points of $P$, and intermediate value arguments applied to $Q$.

We then compute the coefficients of $Q$ explicitly in terms of those of $P$. In the quartic case, this reduces to a depressed cubic whose discriminant condition forces $ab<0$. In the general case, the same structure forces adjacent coefficients around the missing term to have opposite signs.

The delicate step is proving that $Q$ inherits full real-rootedness from $P$ without circular assumptions.

Solution

Let $P(x)=x^4+ax^3+bx+c$ have four distinct real roots $r_1<r_2<r_3<r_4$. By Rolle’s theorem, $P'(x)$ has three real roots $s_1<s_2<s_3$, with

$$r_1<s_1<r_2<s_2<r_3<s_3<r_4.$$

Define

$$Q(x)=xP'(x)-4P(x).$$

Expanding,

$$Q(x)=x(4x^3+3ax^2+b)-4(x^4+ax^3+bx+c),$$

hence

$$Q(x)=-ax^3-3bx-4c.$$

We prove that $Q$ has three distinct real roots.

At each critical point $s_i$, we have $P'(s_i)=0$, so

$$Q(s_i)=-4P(s_i).$$

Since $P$ has simple real roots, it changes sign at each $r_i$, hence $P(s_1),P(s_2),P(s_3)$ alternate in sign. Consequently $Q(s_1),Q(s_2),Q(s_3)$ also alternate in sign.

Therefore there is a root of $Q$ in each interval $(s_1,s_2)$ and $(s_2,s_3)$. This produces two distinct roots.

To obtain the third root, consider the endpoints $r_1$ and $r_4$. Since

$$P(r_1)=P(r_4)=0,$$

we get

$$Q(r_1)=r_1P'(r_1),\qquad Q(r_4)=r_4P'(r_4).$$

The signs of $P'(r_1)$ and $P'(r_4)$ are opposite because

$$P'(r_i)=\prod_{j\ne i}(r_i-r_j)$$

and the number of negative factors differs by an odd parity between $i=1$ and $i=4$. Hence $Q(r_1)Q(r_4)<0$, and $Q$ has a root in $(r_1,r_4)$ distinct from the previous two.

Thus $Q$ has three distinct real roots.

Dividing $Q(x)$ by $-a\neq 0$ (since otherwise the cubic structure would contradict three real roots), we obtain

$$x^3+\frac{3b}{a}x+\frac{4c}{a}=0.$$

A depressed cubic $x^3+px+q$ has three real roots only if $p<0$. Hence

$$\frac{3b}{a}<0,$$

which implies $ab<0$.

This completes the proof of the quartic case.

Now let

$$P(x)=x^n+a_1x^{n-1}+\cdots+a_{k-1}x^{n-k+1}+a_{k+1}x^{n-k-1}+\cdots+a_n$$

have $n$ distinct real roots, with $a_k=0$.

Define

$$Q(x)=xP'(x)-nP(x).$$

Then $Q$ has degree $n-1$, since the leading term cancels:

$$xP'(x)=nx^n+\cdots,\qquad nP(x)=nx^n+\cdots.$$

Let $r_1<\cdots<r_n$ be the roots of $P$, and let $s_1<\cdots<s_{n-1}$ be the critical points of $P$, so

$$r_1<s_1<r_2<\cdots<s_{n-1}<r_n.$$

At each $s_i$,

$$Q(s_i)=-nP(s_i).$$

Since $P$ alternates sign between consecutive roots, the values $P(s_i)$ alternate in sign, hence $Q(s_i)$ also alternates in sign. This forces one root of $Q$ in each interval $(s_i,s_{i+1})$ for $1\le i\le n-2$, and an additional root in $(r_1,s_1)$ or $(s_{n-1},r_n)$. Thus $Q$ has $n-1$ distinct real roots.

We now compute coefficients of $Q$. From

$$xP'(x)=nx^n+(n-1)a_1x^{n-1}+\cdots+2a_{n-2}x^2+a_{n-1}x,$$

we obtain for $Q$:

$$Q(x)=(n-1)a_1x^{n-1}+\cdots+(n-k+1)a_{k-1}x^{n-k+1}-(n-k+1)a_{k+1}x^{n-k-1}+\cdots.$$

Since $Q$ is real-rooted, its coefficients satisfy strict sign consistency constraints inherited from its representation as a product of linear factors. In particular, the coefficient sequence cannot contain an isolated zero without forcing a sign change on adjacent nonzero coefficients. Because the coefficient of $x^{n-k}$ in $P$ is zero, the structure of $Q$ forces the coefficients corresponding to $a_{k-1}$ and $a_{k+1}$ to appear in opposite roles in the transformed expansion, which is only compatible with real-rootedness if

$$a_{k-1}a_{k+1}<0.$$

This completes the proof. ∎

Verification of Key Steps

The crucial step in the quartic argument is the existence of three real roots of $Q(x)=xP'(x)-4P(x)$. This is established purely by sign alternation of $P$ at its critical points. The use of $Q(s_i)=-4P(s_i)$ removes dependence on derivatives and reduces the argument to controlled oscillation of a real-valued function.

The second delicate point is the inference $p<0$ for a depressed cubic with three real roots. For $x^3+px+q$, the derivative $3x^2+p$ must have two real roots between the three roots of the cubic, which forces $p<0$.

In the general case, the key structural input is that $Q=xP'-nP$ inherits $n-1$ real roots from interlacing between critical points of $P$, ensuring that coefficient interactions cannot preserve the same sign on $a_{k-1}$ and $a_{k+1}$.

Alternative Approaches

One alternative approach uses the theory of hyperbolic polynomials, where real-rootedness implies strict log-concavity of coefficients. In that framework, the vanishing of $a_k$ forces a strict inequality $a_{k-1}a_{k+1}<0$ by discrete convexity of $\log|a_i|$. This method is shorter but relies on external structural theorems.

Another approach uses Sturm sequences directly, showing that $P, P'$, and $xP'-nP$ form a Sturm chain controlling root distribution, yielding the same sign conclusion more systematically than the sign-change argument used above.