Kvant Math Problem 186

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Problem

Find all integer solutions, different from 1, of the equation $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$.

V. Slepchenko, 10th-grade student

Exploration

We seek all integer triples $(x,y,z)$, none equal to $1$, satisfying

$$\frac1x+\frac1y+\frac1z=1.$$

Since the equation is symmetric, we may assume

$$|x|\le |y|\le |z|.$$

First consider whether any variable can be negative. If two are negative and one is positive, the left-hand side is at most the positive reciprocal, hence at most $1$, and equality would force the positive variable to be $1$, which is forbidden. If exactly one variable is negative, then the sum of the two positive reciprocals must exceed $1$. Since neither positive variable equals $1$, their reciprocals are at most $1/2$, so the sum is at most $1$. Equality occurs only for $2$ and $2$, leaving no room for the negative term. Thus all variables must be positive.

Now $x\ge2$. Since

$$1=\frac1x+\frac1y+\frac1z\le \frac3x,$$

we get $x\le3$. Hence only $x=2$ or $x=3$ are possible.

If $x=3$, then

$$\frac1y+\frac1z=\frac23.$$

Because $y\ge3$,

$$\frac1y+\frac1z\le \frac23.$$

Equality forces $y=z=3$.

If $x=2$, then

$$\frac1y+\frac1z=\frac12.$$

Multiplying by $2yz$ gives

$$(y-2)(z-2)=4.$$

The positive factor pairs of $4$ are $(1,4)$, $(2,2)$, $(4,1)$, yielding $(y,z)=(3,6),(4,4),(6,3)$.

These seem to be all positive solutions. We must still verify rigorously that no negative solutions exist. That is the step most likely to hide an error.

Problem Understanding

We must determine all integer triples $(x,y,z)$, with each variable different from $1$, satisfying

$$\frac1x+\frac1y+\frac1z=1.$$

This is a Type A problem, a complete classification problem.

The core difficulty is proving that no solution containing negative integers exists. Once positivity is established, simple bounding reduces the problem to two small cases.

The expected answer is the set of permutations of

$$(3,3,3),\qquad (2,4,4),\qquad (2,3,6).$$

The reason is that positive solutions of the Egyptian-fraction equation are heavily constrained: the smallest denominator must be $2$ or $3$, and each case leads to a finite factorization problem.

Proof Architecture

Lemma 1. No solution contains a negative integer. Sketch: examine separately the cases of one negative variable and two negative variables, using that every positive variable is at least $2$.

Lemma 2. Every solution consists of positive integers at least $2$. Sketch: apply Lemma 1 and the condition that no variable equals $1$.

Lemma 3. After reordering so that $2\le x\le y\le z$, one has $x\le3$. Sketch: compare $1/x+1/y+1/z$ with $3/x$.

Lemma 4. If $x=3$, then $(x,y,z)=(3,3,3)$. Sketch: use $y,z\ge3$ and compare with the required sum $2/3$.

Lemma 5. If $x=2$, then $(y-2)(z-2)=4$. Sketch: transform the equation algebraically.

Lemma 6. The factorization of $4$ yields exactly $(y,z)=(3,6)$ or $(4,4)$ up to order. Sketch: list all positive divisor pairs of $4$.

The hardest direction is excluding negative solutions. Lemma 1 is the point most likely to fail under scrutiny if cases are not handled exhaustively.

Solution

Let $(x,y,z)$ be an integer solution of

$$\frac1x+\frac1y+\frac1z=1,$$

with

$$x\ne1,\qquad y\ne1,\qquad z\ne1.$$

We first show that all three integers are positive.

Suppose exactly one of $x,y,z$ is negative. The other two are positive integers different from $1$, hence at least $2$. Their reciprocals are therefore at most $1/2$, so

$$\frac1x+\frac1y+\frac1z < \frac12+\frac12 =1,$$

because the negative reciprocal decreases the sum. This contradicts the equation.

Suppose exactly two of $x,y,z$ are negative. Then the remaining variable is positive. Let it be $z$. Since $z\ne1$, we have $z\ge2$, hence

$$\frac1z\le\frac12.$$

The other two reciprocals are negative, so

$$\frac1x+\frac1y+\frac1z<\frac1z\le\frac12,$$

again contradicting the equation.

All three variables cannot be negative because then the left-hand side would be negative.

Thus every solution consists of positive integers. Since none equals $1$,

$$x,y,z\ge2.$$

Because the equation is symmetric, we may reorder the variables and assume

$$2\le x\le y\le z.$$

Then

$$1=\frac1x+\frac1y+\frac1z\le \frac3x,$$

so

$$x\le3.$$

Hence only two cases are possible.

First, let $x=3$. Then

$$\frac1y+\frac1z=1-\frac13=\frac23.$$

Since $y,z\ge3$,

$$\frac1y+\frac1z\le\frac13+\frac13=\frac23.$$

The required equality can hold only when

$$y=z=3.$$

This yields the solution

$$(3,3,3).$$

Second, let $x=2$. Then

$$\frac1y+\frac1z=\frac12.$$

Multiplying by $2yz$ gives

$$2z+2y=yz.$$

Rearranging,

$$yz-2y-2z=0.$$

Adding $4$ to both sides,

$$(y-2)(z-2)=4.$$

Since $y,z\ge2$, both factors are positive integers. The positive factor pairs of $4$ are

$$(1,4),\qquad (2,2),\qquad (4,1).$$

Therefore

$$(y,z)=(3,6),\ (4,4),\ (6,3).$$

Under the ordering $y\le z$, this gives

$$(y,z)=(3,6)\quad\text{or}\quad(4,4).$$

Hence the ordered solutions with $x\le y\le z$ are

$$(2,3,6),\qquad (2,4,4),\qquad (3,3,3).$$

Because the original equation is symmetric, every permutation of these triples is also a solution, and every solution arises in this way.

Therefore the complete set of integer solutions is

$$\boxed{\text{all permutations of }(3,3,3),\ (2,4,4),\ (2,3,6).}$$

Verification of Key Steps

The exclusion of one negative variable requires more than observing that the positive reciprocals are bounded by $1/2$. If exactly one variable is negative, then the sum of the two positive reciprocals must exceed $1$. For positive integers different from $1$, the largest possible value of such a sum is

$$\frac12+\frac12=1.$$

Since the negative reciprocal is strictly negative, the total sum is strictly less than $1$. This contradiction is decisive.

The deduction $x\le3$ uses the ordering $x\le y\le z$. From this ordering,

$$\frac1y\le\frac1x,\qquad \frac1z\le\frac1x,$$

hence

$$1=\frac1x+\frac1y+\frac1z\le\frac3x.$$

If one forgets to impose the ordering first, the inequality need not hold.

The transformation

$$\frac1y+\frac1z=\frac12 \quad\Longrightarrow\quad (y-2)(z-2)=4$$

is checked directly:

$$\frac1y+\frac1z=\frac12 \Longleftrightarrow 2y+2z=yz \Longleftrightarrow yz-2y-2z+4=4 \Longleftrightarrow (y-2)(z-2)=4.$$

Since $y,z\ge2$, both factors are positive, so no factor pairs are missed.

Alternative Approaches

A different approach begins with

$$\frac1x+\frac1y+\frac1z=1$$

and assumes $x\le y\le z$. Multiplying by $xyz$ gives

$$xy+xz+yz=xyz.$$

Treating this as a quadratic equation in $z$ yields

$$z=\frac{xy}{xy-x-y}.$$

The denominator must be positive, so $(x-1)(y-1)>1$. Since $x\le y$, one obtains $x\le3$ by estimating the denominator. The cases $x=2$ and $x=3$ are then handled exactly as above.

The main approach is preferable because it isolates positivity first and then reduces the problem to a short factorization identity, avoiding a more complicated divisibility analysis.