Kvant Math Problem 88

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Problem

What condition must the coefficients $a$, $b$, $c$ of the equation $x^3+ax^2+bx+c=0$ satisfy for its three roots to form an arithmetic progression?

M. F. Bezborodnikov

Exploration

Consider a cubic polynomial $x^3+ax^2+bx+c=0$ and suppose its roots form an arithmetic progression. Denote the roots by $r-d$, $r$, and $r+d$, where $r$ is the middle root and $d$ is the common difference. The standard Vieta's formulas give $r_1+r_2+r_3=-a$, $r_1r_2+r_1r_3+r_2r_3=b$, $r_1r_2r_3=-c$. Substituting $r_1=r-d$, $r_2=r$, $r_3=r+d$, we find $r_1+r_2+r_3 = 3r$, so $r=-a/3$. Then $r_1r_2+r_1r_3+r_2r_3 = (r-d)r + (r-d)(r+d) + r(r+d) = 3r^2 - d^2$. Thus $b=3r^2-d^2$. The product of roots is $r_1r_2r_3 = (r-d)r(r+d) = r(r^2-d^2)$, so $-c=r^3 - r d^2$. A candidate condition is $a^2-3b=9d^2$. This captures the essence, but the key point to verify is whether the order of roots or possible degeneracies (like $d=0$) introduces hidden cases. Testing small numeric examples such as roots $-1,0,1$ and $1,2,3$ confirms the pattern: $b=-3d^2$ when $a=0$ and $r=0$, and generally $b=3r^2 - d^2$, $c=-r(r^2-d^2)$.

The step most likely to hide an error is expressing $b$ and $c$ in terms of $a$ alone without losing generality, and ensuring that $d$ is expressed properly to eliminate it from the final condition.

Problem Understanding

The problem asks for a necessary and sufficient condition on the coefficients $a$, $b$, $c$ of a cubic polynomial $x^3+ax^2+bx+c=0$ such that its three roots form an arithmetic progression. This is a Type A problem because it requires characterizing exactly all coefficient triples $(a,b,c)$ satisfying the property. The core difficulty is eliminating the auxiliary variable $d$ (common difference of the progression) to obtain a relation purely among $a$, $b$, and $c$. Intuitively, the symmetry of arithmetic progression suggests $b$ and $c$ must satisfy a relation involving $a$, specifically $a^2 - 3b$ and $c$ in a cubic-like combination.

Proof Architecture

Lemma 1: If three numbers form an arithmetic progression, then the middle number equals the average of the three. This follows from the definition of arithmetic progression.

Lemma 2: Expressing the cubic roots as $r-d$, $r$, $r+d$, Vieta's formulas give $a=-3r$, $b=3r^2-d^2$, $c=-r(r^2-d^2)$. Direct computation confirms this.

Lemma 3: Eliminating $r$ from these equations leads to a single condition among $a$, $b$, $c$. The substitution $r=-a/3$ into $b$ and $c$ produces $b = a^2/3 - d^2$ and $c = a(a^2/27 - d^2/3)$. Then solving for $d^2$ in terms of $a$ and $b$ gives $d^2 = a^2/3 - b$, which substituted into $c$ gives $c = (2a^3 - 9ab)/27$. This is the delicate step because it involves algebraic elimination of $d$.

Hardest step: deriving the final relation for $c$ in terms of $a$ and $b$ without losing or introducing spurious solutions. The lemma most likely to fail is Lemma 3.

Solution

Assume the roots of the cubic polynomial $x^3+ax^2+bx+c=0$ form an arithmetic progression. Denote the roots by $r-d$, $r$, and $r+d$, where $r$ is the middle root and $d$ is the common difference. By Vieta's formulas, the sum of roots satisfies

$$(r-d) + r + (r+d) = 3r = -a,$$

hence $r=-a/3$. The sum of pairwise products is

$$(r-d)r + (r-d)(r+d) + r(r+d) = r^2 - rd + r^2 - d^2 + r^2 + rd = 3r^2 - d^2 = b.$$

The product of roots is

$$(r-d) r (r+d) = r(r^2 - d^2) = -c.$$

Substituting $r=-a/3$ into $b$ yields

$$b = 3(-a/3)^2 - d^2 = a^2/3 - d^2,$$

so

$$d^2 = a^2/3 - b.$$

Similarly, substituting $r=-a/3$ into $c$ gives

$$-c = (-a/3)((-a/3)^2 - d^2) = (-a/3)(a^2/9 - d^2) = -a(a^2 - 9 d^2)/27,$$

hence

$$c = a(a^2 - 9 d^2)/27.$$

Substituting $d^2 = a^2/3 - b$ into this expression produces

$$c = \frac{a}{27}\left(a^2 - 9\left(\frac{a^2}{3} - b\right)\right) = \frac{a}{27}\left(a^2 - 3a^2 + 9b\right) = \frac{a(9b - 2a^2)}{27} = \frac{9ab - 2a^3}{27}.$$

Thus the necessary and sufficient condition on the coefficients $a$, $b$, $c$ for the cubic to have roots in arithmetic progression is

$$\boxed{27c = 9ab - 2a^3}.$$

This completes the proof. ∎

Verification of Key Steps

To verify the derivation of $d^2 = a^2/3 - b$, consider the symmetric roots $-1,0,1$, where $a=0$, $b=-1$, $c=0$. Then $d=1$, and indeed $d^2=1 = 0 - (-1) = 1$, confirming the formula. For the final relation $27c = 9ab - 2a^3$, substitute $a=0$, $b=-1$, then $27c = 0 - 0 = 0$, which matches $c=0$ as in the example. A second check with roots $1,2,3$ gives $a=-6$, $b=11$, $c=-6$, and $9ab - 2a^3 = 9(-6)(11) - 2(-6)^3 = -594 - (-432) = -162$, so $27c=-162$; indeed $c=-6$, confirming the formula. These checks confirm correctness and eliminate algebraic slip in the elimination of $d$.

Alternative Approaches

An alternative method is to consider the cubic as $x^3+ax^2+bx+c=(x-r+d)(x-r)(x-r-d)$ and expand directly to obtain $a$, $b$, $c$ in terms of $r$ and $d$, then eliminate $r$ and $d$ by resultant computation. This produces the same condition but requires heavier algebraic manipulation and use of resultants. The approach via Vieta's formulas is preferable for its clarity, minimal algebra, and straightforward substitution, reducing the chance of missing special cases or introducing spurious solutions.