Kvant Math Problem 357
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m56s
Source on kvant.digital
Problem
Prove that if $x+\dfrac1y=y+\dfrac1z=z+\dfrac1x$, then $x=y=z$ or $x^2y^2z^2=1$.
Exploration
Let
$$x+\frac1y=y+\frac1z=z+\frac1x=k.$$
Subtracting consecutive equalities gives
$$x-y=\frac1z-\frac1y=\frac{y-z}{yz},$$
$$y-z=\frac1x-\frac1z=\frac{z-x}{xz},$$
$$z-x=\frac1y-\frac1x=\frac{x-y}{xy}.$$
These relations connect the differences $x-y$, $y-z$, $z-x$ with one another. If all three differences are zero, then $x=y=z$, which is one of the required conclusions.
The interesting case is when not all differences vanish. Then at least one of them is nonzero. From the first relation,
$$yz(x-y)=y-z.$$
The second gives
$$xz(y-z)=z-x,$$
and the third gives
$$xy(z-x)=x-y.$$
Multiplying these three equalities seems promising because the differences appear cyclically. Doing so yields
$$x^2y^2z^2(x-y)(y-z)(z-x) =(y-z)(z-x)(x-y).$$
If the product of differences is nonzero, cancellation gives
$$x^2y^2z^2=1.$$
The only point requiring care is the cancellation. One must first separate the case when some difference is zero. If one difference is zero, the original equations force all differences to be zero, giving $x=y=z$. Thus the nontrivial case indeed has all three differences nonzero, and cancellation is legitimate.
The crucial step most likely to hide an error is proving that if one pair of variables is equal, then all three are equal.
Problem Understanding
We are given
$$x+\frac1y=y+\frac1z=z+\frac1x,$$
and must prove that either all three variables are equal or
$$x^2y^2z^2=1.$$
This is a Type B problem, a pure proof.
The core difficulty is handling the possibility that some of the differences $x-y$, $y-z$, $z-x$ vanish. The algebraic manipulation leading to $x^2y^2z^2=1$ requires cancellation of these differences, so the equal-variable cases must be treated separately and rigorously.
Proof Architecture
First show that if any two of $x,y,z$ are equal, then all three are equal; substituting into the given equations yields the third equality immediately.
Next derive the identities
$$yz(x-y)=y-z,\qquad xz(y-z)=z-x,\qquad xy(z-x)=x-y$$
by subtracting consecutive given equalities and clearing denominators.
Then consider the case $x\ne y$, $y\ne z$, $z\ne x$; multiply the three identities together.
After multiplication, cancel the nonzero factor $(x-y)(y-z)(z-x)$ and obtain
$$x^2y^2z^2=1.$$
The hardest direction is justifying the cancellation. The lemma most likely to fail under scrutiny is the claim that the vanishing of one difference forces all three variables to be equal.
Solution
Let
$$x+\frac1y=y+\frac1z=z+\frac1x.$$
We first examine the possibility that two variables are equal.
Assume $x=y$. Comparing the first two expressions,
$$x+\frac1x=x+\frac1z,$$
hence
$$\frac1x=\frac1z.$$
Since the given expressions are defined, $x$ and $z$ are nonzero, so $x=z$. Therefore $x=y=z$.
The same argument works if $y=z$ or $z=x$. Consequently, if any two of the variables are equal, then
$$x=y=z.$$
It remains to consider the case
$$x\ne y,\qquad y\ne z,\qquad z\ne x.$$
Subtracting
$$y+\frac1z$$
from
$$x+\frac1y$$
gives
$$x-y=\frac1z-\frac1y=\frac{y-z}{yz},$$
hence
$$yz(x-y)=y-z.$$
Similarly,
$$y-z=\frac1x-\frac1z=\frac{z-x}{xz},$$
so
$$xz(y-z)=z-x.$$
Finally,
$$z-x=\frac1y-\frac1x=\frac{x-y}{xy},$$
and therefore
$$xy(z-x)=x-y.$$
Multiplying these three identities yields
$$\bigl(yz(x-y)\bigr)\bigl(xz(y-z)\bigr)\bigl(xy(z-x)\bigr) =(y-z)(z-x)(x-y).$$
After collecting factors on the left,
$$x^2y^2z^2,(x-y)(y-z)(z-x) =(x-y)(y-z)(z-x).$$
Because we are in the case
$$x\ne y,\qquad y\ne z,\qquad z\ne x,$$
the factor
$$(x-y)(y-z)(z-x)$$
is nonzero. Cancelling it gives
$$x^2y^2z^2=1.$$
Thus either $x=y=z$, or
$$x^2y^2z^2=1.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the claim that one equality of variables forces all three variables to be equal. Suppose $x=y$. Then the common value of the given expressions satisfies
$$x+\frac1x=x+\frac1z.$$
Subtracting $x$ gives
$$\frac1x=\frac1z.$$
Because reciprocals are defined, $x,z\ne0$, hence $x=z$. No additional assumptions are needed. The same reasoning works cyclically.
The second delicate step is the derivation
$$yz(x-y)=y-z.$$
Starting from
$$x-y=\frac1z-\frac1y,$$
the right-hand side equals
$$\frac{y-z}{yz}.$$
Multiplying by $yz$ is legitimate because $y$ and $z$ are nonzero, as required by the original equations. This yields the identity exactly.
The third delicate step is cancellation after multiplication. Cancellation is valid only when
$$(x-y)(y-z)(z-x)\ne0.$$
The proof isolates the equal-variable case first. Once that case is excluded, each factor is nonzero, and the cancellation is fully justified.
Alternative Approaches
Introduce the common value
$$k=x+\frac1y=y+\frac1z=z+\frac1x.$$
Then
$$x-y=\frac{y-z}{yz},\qquad y-z=\frac{z-x}{xz},\qquad z-x=\frac{x-y}{xy}.$$
These relations form a homogeneous linear system in the differences. If the differences are not all zero, multiplying the three equations immediately gives
$$(xyz)^2=1.$$
If all differences vanish, then $x=y=z$.
This approach is essentially the same algebra packaged in terms of the common value $k$. The main proof is preferable because it separates the equal-variable case explicitly before any cancellation and makes every algebraic step transparent.