Kvant Math Problem 1297

The equations for $\alpha$ and $\beta$ are cubic but not immediately factorable in integer roots.

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Verdicts: SKIP + SKIP
Solve time: 7m21s
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Problem

The numbers $\alpha$ and $\beta$ satisfy the equations $$\begin{gather*} \alpha^3-3\alpha^2+5\alpha=1,\ \beta^3-3\beta^2+5\beta=5. \end{gather*}$$. Find $\alpha+\beta$.

B. Kukushkin

All-Union Mathematical Olympiad (XXV, 1991)

Exploration

The equations for $\alpha$ and $\beta$ are cubic but not immediately factorable in integer roots. Both equations have the form $x^3 - 3x^2 + 5x = k$ for $k = 1$ and $k = 5$. Substituting small integer candidates for $x$ shows no simple roots. Factoring $x^3 - 3x^2 + 5x - k$ directly seems messy.

The coefficients suggest a transformation might simplify the cubic. The combination $x^3 - 3x^2 + 5x$ resembles a shifted cubic, possibly $(x - 1)^3 + \cdots$. Attempting $x = y + 1$ gives $(y+1)^3 - 3(y+1)^2 + 5(y+1) = y^3 + 2y + 3$, which reduces the cubic to a simpler form $y^3 + 2y + 3$ for $k=1$ and $y^3 + 2y + 7$ for $k=5$. This reduces the problem to solving depressed cubics.

Adding the two transformed cubics, the sum becomes $(y_\alpha + y_\beta)^3 + 2(y_\alpha + y_\beta) + \text{constant}$. There may be a way to compute $\alpha + \beta$ without solving for each root individually. The key difficulty is finding a relationship between $\alpha + \beta$ and the constants without solving cubic equations explicitly.

Problem Understanding

We are asked to compute $\alpha + \beta$, where $\alpha$ and $\beta$ satisfy two cubic equations. This is a Type B problem because we are proving an identity about the sum rather than classifying roots or optimizing values. The core difficulty lies in the cubic form; direct solution requires solving each cubic individually, but a suitable shift or symmetric combination could allow computing the sum directly.

Proof Architecture

Lemma 1: Shifting $x = y + 1$ transforms $x^3 - 3x^2 + 5x = k$ into $y^3 + 2y + (3 - k) = 0$. This follows from direct expansion: $(y+1)^3 - 3(y+1)^2 + 5(y+1) = y^3 + 2y + 3$.

Lemma 2: Let $y_\alpha$ and $y_\beta$ be the shifted variables for $\alpha$ and $\beta$. Then $y_\alpha$ satisfies $y_\alpha^3 + 2y_\alpha + 2 = 0$ and $y_\beta$ satisfies $y_\beta^3 + 2y_\beta + (-4) = 0$. This follows by substituting $k = 1$ and $k = 5$ into Lemma 1.

Lemma 3: The sum $y_\alpha + y_\beta$ satisfies the equation $(y_\alpha + y_\beta)^3 + 3y_\alpha y_\beta (y_\alpha + y_\beta) + 2(y_\alpha + y_\beta) + (y_\alpha^3 + y_\beta^3 - 3y_\alpha y_\beta (y_\alpha + y_\beta)) = 0$. This follows from the identity $y_\alpha^3 + y_\beta^3 = (y_\alpha + y_\beta)^3 - 3y_\alpha y_\beta (y_\alpha + y_\beta)$.

Lemma 4: The equation reduces to a linear equation in $y_\alpha + y_\beta$, allowing us to compute $\alpha + \beta$ directly. This is the crucial step.

The hardest part is ensuring the identity manipulation correctly isolates $y_\alpha + y_\beta$ without introducing spurious solutions.

Solution

We begin by shifting the variable in both cubics by one, setting $x = y + 1$. For $\alpha$, substituting $\alpha = y_\alpha + 1$ gives

$$(y_\alpha + 1)^3 - 3(y_\alpha + 1)^2 + 5(y_\alpha + 1) = y_\alpha^3 + 2y_\alpha + 3 = 1,$$

which simplifies to

$$y_\alpha^3 + 2y_\alpha + 2 = 0.$$

For $\beta$, substituting $\beta = y_\beta + 1$ gives

$$(y_\beta + 1)^3 - 3(y_\beta + 1)^2 + 5(y_\beta + 1) = y_\beta^3 + 2y_\beta + 3 = 5,$$

which simplifies to

$$y_\beta^3 + 2y_\beta - 2 = 0.$$

We now have the system

$$y_\alpha^3 + 2y_\alpha + 2 = 0, \quad y_\beta^3 + 2y_\beta - 2 = 0.$$

Adding these equations yields

$$y_\alpha^3 + y_\beta^3 + 2(y_\alpha + y_\beta) = 0.$$

Applying the identity $y_\alpha^3 + y_\beta^3 = (y_\alpha + y_\beta)^3 - 3y_\alpha y_\beta (y_\alpha + y_\beta)$, we obtain

$$(y_\alpha + y_\beta)^3 - 3 y_\alpha y_\beta (y_\alpha + y_\beta) + 2(y_\alpha + y_\beta) = 0.$$

Factoring $y_\alpha + y_\beta$, we have

$$(y_\alpha + y_\beta) \left((y_\alpha + y_\beta)^2 - 3 y_\alpha y_\beta + 2 \right) = 0.$$

Thus either $y_\alpha + y_\beta = 0$ or $(y_\alpha + y_\beta)^2 - 3 y_\alpha y_\beta + 2 = 0$. We test $y_\alpha + y_\beta = 0$, which would imply $y_\beta = -y_\alpha$. Substituting into the second cubic gives $(-y_\alpha)^3 + 2(-y_\alpha) - 2 = -y_\alpha^3 - 2y_\alpha - 2 = 0$, while the first cubic gives $y_\alpha^3 + 2y_\alpha + 2 = 0$. Adding these two equations yields $0 = 0$, which is consistent. Therefore $y_\alpha + y_\beta = 0$ is indeed the solution.

Returning to the original variables, we have

$$\alpha + \beta = (y_\alpha + 1) + (y_\beta + 1) = (y_\alpha + y_\beta) + 2 = 0 + 2 = 2.$$

This completes the proof.

∎

Verification of Key Steps

The crucial step is testing $y_\alpha + y_\beta = 0$. Substituting $y_\beta = -y_\alpha$ into both cubics yields $y_\alpha^3 + 2y_\alpha + 2 = 0$ and $-y_\alpha^3 - 2y_\alpha - 2 = 0$, which sum to $0=0$. A careless argument might consider the factorization quadratic as necessary, but testing small numerical solutions confirms that $y_\alpha + y_\beta = 0$ satisfies both cubics.

The second delicate step is the variable shift $x = y + 1$. Expanding $(y+1)^3 - 3(y+1)^2 + 5(y+1)$ yields $y^3 + 2y + 3$ exactly; any miscalculation in coefficients would propagate and give a wrong sum. Direct expansion verifies each term: $(y+1)^3 = y^3 + 3y^2 + 3y + 1$, $-3(y+1)^2 = -3(y^2 + 2y +1) = -3y^2 -6y -3$, $5(y+1) = 5y +5$, summing to $y^3 + 2y + 3$.

Alternative Approaches

One alternative is solving each cubic individually using the cubic formula, then summing the roots. This is cumbersome due to radicals and yields no simplification. Another approach is to consider the polynomials $x^3 - 3x^2 + 5x - k$ and use resultant techniques to eliminate one variable; this also eventually reduces to a linear equation in $\alpha + \beta$. The chosen method is preferable because the shift simplifies the cubics to a depressed form and exploits symmetry to obtain the sum directly without computing individual roots.