Kvant Math Problem 703
The first equation links three expressions of the form $t + \frac{1}{t}$ multiplied by constants 3, 4, and 5.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m04s
Source on kvant.digital
Problem
Solve the system of equations $$\left{\begin{array}{l} 3\left(x+\dfrac{1}{x}\right)=4\left(y+\dfrac{1}{y}\right)=5\left(z+\dfrac{1}{z}\right),\ xy+yz+zx=1. \end{array}\right.$$
A. Fyodorov
Exploration
The first equation links three expressions of the form $t + \frac{1}{t}$ multiplied by constants 3, 4, and 5. Denoting $u = x + \frac{1}{x}$, $v = y + \frac{1}{y}$, $w = z + \frac{1}{z}$, the system reduces to $3u = 4v = 5w$. This suggests the ratios $\frac{u}{v} = \frac{4}{3}$, $\frac{u}{w} = \frac{5}{3}$, etc. Each equation $t + \frac{1}{t} = \text{constant}$ can be rewritten as a quadratic in $t$, giving two roots per variable.
The second equation, $xy + yz + zx = 1$, is symmetric in $x, y, z$. A natural approach is to express $x, y, z$ through $u, v, w$, then impose the linear relation among the pairwise products. Small integers or simple fractions for $x, y, z$ could yield rational sums of products, given the integer constants 3, 4, 5. It is likely that the solution set is finite and may involve reciprocals like $1$ or $-1$.
A plausible key insight is that $x, y, z$ must satisfy $x + 1/x = 4/3 y + 1/y = 5/3 z + 1/z$; the ratios of the constants might suggest a proportionality among the variables themselves, possibly $x : y : z = 1 : 1 : 1$ or $x : y : z = 1 : 2 : 3$, but the pairwise product sum $xy + yz + zx = 1$ restricts these guesses. The most delicate step will be ensuring no solutions are lost when solving each quadratic for $x, y, z$.
Problem Understanding
The problem asks to find all real triples $(x, y, z)$ satisfying both $3(x + \frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z + \frac{1}{z})$ and $xy + yz + zx = 1$. This is a Type A problem: a classification problem, requiring both verification that any claimed solution works and proof that no other solutions exist.
The core difficulty lies in handling the first equation, which links three nonlinear expressions, while simultaneously satisfying a symmetric quadratic constraint on the products. The answer likely consists of a small finite set of rational or simple algebraic numbers, derived from solving the quadratic forms $t + 1/t = \text{constant}$ and enforcing $xy + yz + zx = 1$.
Heuristically, one expects the solution to involve numbers such that $x = y = z$ or numbers with small integer ratios, because the sum of reciprocals terms must be rationally proportional.
Proof Architecture
Lemma 1: If $3(x + 1/x) = 4(y + 1/y) = 5(z + 1/z)$, then $x + 1/x$, $y + 1/y$, and $z + 1/z$ are positive and proportional. Sketch: Denote the common value $k = 3(x + 1/x) = 4(y + 1/y) = 5(z + 1/z)$; then $x + 1/x = k/3$, $y + 1/y = k/4$, $z + 1/z = k/5$.
Lemma 2: For each positive $k$, the equation $t + 1/t = k/c$ has at most two real solutions, one larger than 1 and one between 0 and 1. Sketch: The function $f(t) = t + 1/t$ is strictly decreasing on $(0,1]$ and strictly increasing on $[1, \infty)$, attaining its minimum at $t = 1$, so each $k/c \ge 2$ yields two positive solutions.
Lemma 3: The equation $xy + yz + zx = 1$ restricts the signs and approximate sizes of $x, y, z$. Sketch: If all variables are positive, each quadratic solution must produce numbers whose pairwise products sum to 1. Negative solutions lead to a sum of products less than zero, which is impossible.
Lemma 4: The only solution triples arise when $x, y, z$ are all positive. Sketch: Checking sign patterns shows that any negative component violates the product sum equality.
Lemma 5: Solving explicitly using $x = k/3 - 1/x$, $y = k/4 - 1/y$, $z = k/5 - 1/z$, then substituting into $xy + yz + zx = 1$ and solving for $k$ yields finitely many candidates. Sketch: Express each variable as the positive root of its quadratic and reduce the product sum to a quartic in $k$, then find the unique solution compatible with all positivity constraints.
The hardest step is Lemma 5, because it requires explicit computation and verification that no additional solutions exist for negative roots.
Solution
Let $k$ denote the common value in the first equation, so that $k = 3(x + 1/x) = 4(y + 1/y) = 5(z + 1/z)$. Then $x + 1/x = k/3$, $y + 1/y = k/4$, and $z + 1/z = k/5$. Each equation $t + 1/t = m$ is quadratic: $t^2 - m t + 1 = 0$. Therefore, $x$ satisfies $x^2 - (k/3)x + 1 = 0$, $y^2 - (k/4)y + 1 = 0$, $z^2 - (k/5)z + 1 = 0$.
For each quadratic, the discriminants are $\Delta_x = (k/3)^2 - 4$, $\Delta_y = (k/4)^2 - 4$, $\Delta_z = (k/5)^2 - 4$. For real solutions, each discriminant must be nonnegative. Thus $k/3 \ge 2 \implies k \ge 6$, $k/4 \ge 2 \implies k \ge 8$, $k/5 \ge 2 \implies k \ge 10$. All three inequalities are satisfied only if $k \ge 10$.
Denote the positive roots by $x = \frac{k/3 + \sqrt{(k/3)^2 - 4}}{2}$, $y = \frac{k/4 + \sqrt{(k/4)^2 - 4}}{2}$, $z = \frac{k/5 + \sqrt{(k/5)^2 - 4}}{2}$. The pairwise product sum is
$$xy + yz + zx = 1.$$
Substituting the expressions for $x, y, z$ gives a complicated function of $k$, but we can attempt a rational solution by testing integer values of $k$ starting from the minimal feasible $k = 10$.
For $k = 10$, $x = \frac{10/3 + \sqrt{100/9 - 4}}{2} = \frac{10/3 + \sqrt{64/9}}{2} = \frac{10/3 + 8/3}{2} = 3$, $y = \frac{10/4 + \sqrt{100/16 - 4}}{2} = \frac{5/2 + \sqrt{25/4 - 4}}{2} = \frac{5/2 + 3/2}{2} = 2$, $z = \frac{10/5 + \sqrt{100/25 - 4}}{2} = \frac{2 + \sqrt{4 - 4}}{2} = 1$.
Compute $xy + yz + zx = 3\cdot 2 + 2\cdot 1 + 3\cdot 1 = 6 + 2 + 3 = 11 \neq 1$.
Try $k = 12$. Then $x = \frac{12/3 + \sqrt{16 - 4}}{2} = \frac{4 + \sqrt{12}}{2} = 2 + \sqrt{3}$, $y = \frac{12/4 + \sqrt{9 - 4}}{2} = \frac{3 + \sqrt{5}}{2}$, $z = \frac{12/5 + \sqrt{(12/5)^2 - 4}}{2} = \frac{12/5 + \sqrt{144/25 - 4}}{2} = \frac{12/5 + \sqrt{44/25}}{2} = \frac{12/5 + 2\sqrt{11}/5}{2} = \frac{6 + \sqrt{11}}{5}$. The sum $xy + yz + zx$ is clearly larger than 1.
To satisfy $xy + yz + zx = 1$, consider smaller $k$ corresponding to negative roots. Try negative roots, taking $x = \frac{k/3 - \sqrt{(k/3)^2 - 4}}{2}$, $y = \frac{k/