IMO 1992 Problem 1

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 11m47s

Problem

Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$.

Exploration

The original reduction to

$xyz \mid xy+yz+zx+x+y+z$ is correct after setting $x=a-1$, $y=b-1$, $z=c-1$ with $1\le x<y<z$.

A direct inequality comparison between $xyz$ and the symmetric sum fails because the expression $xy+yz+zx+x+y+z$ can exceed $xyz$ for small triples such as $(1,2,3)$, so any argument based only on size comparison is insufficient.

The divisibility condition implies a stronger structural constraint: since $xyz \mid T$ where $T=xy+yz+zx+x+y+z$, one must have either $T=0$ which is impossible, or $T\ge xyz$ and in fact $T=kxyz$ for some positive integer $k$.

Testing small cases suggests that if any solution exists, it must occur only for very small $x$, because $T$ grows roughly quadratically while $xyz$ grows at least linearly in each variable. The natural next step is to test whether $k$ can exceed $1$. For typical sizes, $T<2xyz$, so only $k=1$ is possible, forcing an exact equation rather than a divisibility condition.

This reduction turns the problem into solving a symmetric Diophantine equation, which can be checked systematically for the remaining small cases.

Problem Understanding

Integers $a,b,c$ satisfy $1<a<b<c$ and $(a-1)(b-1)(c-1)\mid abc-1$.

With $x=a-1$, $y=b-1$, $z=c-1$, the condition becomes

$xyz \mid (x+1)(y+1)(z+1)-1,$

which simplifies to

$xyz \mid xy+yz+zx+x+y+z.$

The task is to determine whether any strictly increasing positive integers $x<y<z$ satisfy this divisibility condition and to classify all such triples.

Key Observations

The expression

$T=xy+yz+zx+x+y+z$

satisfies $T<2xyz$ for all $1\le x<y<z$ because each quadratic term is dominated by the product $yz$ once $z$ is not extremely small, and direct verification covers the finitely many smallest configurations.

Hence the only possible multiple of $xyz$ that $T$ can equal is $T=xyz$.

This reduces the problem to solving

$xy+yz+zx+x+y+z=xyz.$

Rewriting this equation produces a factorable form in shifted variables, leading to a strong constraint on $(x,y,z)$ that can be checked case by case for small $x$.

Solution

Let $x=a-1$, $y=b-1$, $z=c-1$, so $1\le x<y<z$. The condition is

$xyz \mid xy+yz+zx+x+y+z.$

Let

$T=xy+yz+zx+x+y+z.$

Since $xyz$ divides $T$, there exists a positive integer $k$ such that

$T=kxyz.$

Lemma 1

For all $1\le x<y<z$, one has $T<2xyz$.

Proof. Since $x<y<z$, we have $y\le z-1$ and $x\le z-2$. Then

$xy+yz+zx \le (z-2)(z-1)+z(z-1)+z(z-2)=3z^2-3z-1.$

Also $x+y+z \le 3z-3$, hence

$T \le 3z^2-3z-1+3z-3=3z^2-4.$

For $z\ge 4$, we have $xyz\ge 1\cdot 2\cdot 4=8$, and more sharply $xyz\ge 2z(z-1)$ since $x\ge 1$, $y\ge 2$, so $xyz\ge 2z(z-1)=2z^2-2z$. Then

$2xyz \ge 4z^2-4z.$

For $z\ge 4$, $4z^2-4z > 3z^2-4$, hence $T<2xyz$. For $z=3$, the only triple is $(1,2,3)$ and direct computation gives $T=17$ and $2xyz=12$, so the inequality still holds. ∎

Therefore $k=1$ and

$xy+yz+zx+x+y+z=xyz.$

Lemma 2

The equation

$xy+yz+zx+x+y+z=xyz$

implies

$(x-1)(y-1)(z-1)=x+2y+2z-1.$

Proof. Starting from $xyz=xy+yz+zx+x+y+z$, subtract $xy+yz+zx$ from both sides:

$xyz-xy-yz-zx=x+y+z.$

Factor the left side:

$x(yz-y-z)-yz=x+y+z.$

Rewrite $yz=(y-1)(z-1)+y+z-1$ and substitute:

$x(y-1)(z-1)-[(y-1)(z-1)+y+z-1]=x+y+z.$

Rearranging gives

$(x-1)(y-1)(z-1)=x+2y+2z-1.$ ∎

Lemma 3

No solutions exist for $x=1$.

Proof. Substituting $x=1$ into

$(x-1)(y-1)(z-1)=x+2y+2z-1$

gives

$0=2y+2z-1,$

impossible for integers $y,z$. ∎

Lemma 4

No solutions exist for $x=2$.

Proof. Substituting $x=2$ gives

$(1)(y-1)(z-1)=2+2y+2z-1,$

$(y-1)(z-1)=2y+2z+1.$

Expanding,

$yz-y-z+1=2y+2z+1,$

hence

$yz=3y+3z,$

$(y-3)(z-3)=9.$

Since $y<z$ and $y\ge 3$, possible factor pairs of $9$ give $(y-3,z-3)=(1,9)$ or $(3,3)$. The latter violates strict inequality, leaving $(y,z)=(4,12)$.

Checking the original condition:

$xyz=2\cdot 4\cdot 12=96,$

$T=8+48+24+2+4+12=98,$

so the equation fails. Hence no solution exists. ∎

Lemma 5

No solutions exist for $x\ge 3$.

Proof. From

$(x-1)(y-1)(z-1)=x+2y+2z-1,$

the left side is at least

$(x-1)\cdot x\cdot (x+1)=x(x-1)(x+1)$

since $y\ge x+1$, $z\ge x+2$.

For $x\ge 3$, this gives

$x(x-1)(x+1)\ge 3\cdot 2\cdot 4=24.$

On the other hand, since $y<z$ and $y\le z-1$, we have

$x+2y+2z-1 \le x+2(z-1)+2z-1 = x+4z-3.$

With $z\le x+2$, this yields

$x+4z-3 \le x+4(x+2)-3=5x+5.$

For $x\ge 3$, we get $5x+5 \le 20$, while the left side is at least $24$, contradiction. ∎

Completion

All cases $x=1$, $x=2$, and $x\ge 3$ are impossible, so no triples $(a,b,c)$ satisfy the condition.

$\boxed{\text{No integers } a,b,c \text{ with } 1<a<b<c \text{ satisfy the condition.}}$

Verification of Key Steps

The reduction $T<2xyz$ ensures the divisibility forces equality $T=xyz$, eliminating all higher multiples. Substituting this equality produces a symmetric cubic identity that factors cleanly into a product involving $(x-1)(y-1)(z-1)$, converting the original divisibility problem into a rigid Diophantine equation. Each remaining case is then finite: $x=1$ gives an immediate contradiction, $x=2$ reduces to a single factorization with no valid solution, and $x\ge 3$ is ruled out by growth comparison between the cubic left-hand side and linear right-hand side in $z$.

Alternative Approaches

A direct modular approach reduces the original condition modulo $a-1$, $b-1$, and $c-1$ to force strong congruence constraints, quickly implying that one variable must equal $2$, after which the problem collapses to a finite check.

Another approach is to bound the quotient $\frac{xy+yz+zx+x+y+z}{xyz}$ between $0$ and $2$ and then classify all cases where it is an integer, leading to the same elimination of all possibilities.