IMO 1998 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 14m26s

Problem

Determine all pairs $(a, b)$ of positive integers such that $ab^{2} + b + 7$ divides $a^{2}b + a + b$.

Exploration

The divisibility condition is $ab^{2}+b+7 \mid a^{2}b+a+b$. Writing $D=ab^{2}+b+7$ and $N=a^{2}b+a+b$, the combination $bN-aD$ eliminates the highest degree terms and gives $b^{2}-7a$. This shows $D \mid (b^{2}-7a)$.

Substituting small values suggests that most solutions occur only when $b$ is small or when a rigid relation between $a$ and $b$ holds. The case $b=1$ produces a small finite set. The case $b=2$ appears empty. For $b\ge 3$, the divisor grows like $ab^{2}$ while the numerator grows like $a^{2}b$, suggesting that $a$ must remain small unless a special cancellation occurs.

Testing the structured family $b^{2}=7a$ produces consistent divisibility, and no other stable algebraic patterns appear under small substitutions.

These observations indicate that the correct solution consists of one parametric family and finitely many exceptional cases coming only from small $b$.

Problem Understanding

Positive integers $a,b$ satisfy

$ab^{2}+b+7 \mid a^{2}b+a+b.$

The task is to determine all such pairs. A complete solution must first construct all valid pairs and then prove that no others exist.

Key Observations

Let

$D=ab^{2}+b+7,\quad N=a^{2}b+a+b.$

The identity

$bN-aD=b^{2}-7a$

implies

$D \mid (b^{2}-7a).$

Hence there exists an integer $t$ such that

$b^{2}-7a=t(ab^{2}+b+7).$

Rearranging gives

$a(7+tb^{2})=b^{2}-tb-7t,$

$a=\frac{b^{2}-tb-7t}{7+tb^{2}}.$

This representation shows that $t=0$ produces a structured infinite family, while $t\ne 0$ forces strong arithmetic restrictions that limit $b$ to small values.

Solution

Case 1: $t=0$

If $t=0$, then $b^{2}=7a$. Since $7$ is prime, $7 \mid b$, so write $b=7m$. Then $a=7m^{2}$.

Substituting,

$D=7m^{2}\cdot 49m^{2}+7m+7=343m^{4}+7m+7,$

$N=49m^{4}\cdot 7m+7m^{2}+7m=343m^{5}+7m^{2}+7m.$

Factoring yields

$N=mD,$

so $D \mid N$ for all $m\ge 1$. This gives the family

$(a,b)=(7m^{2},7m).$

Case 2: $t\ne 0$

From

$|b^{2}-7a|=|t|D \ge D \ge ab^{2},$

we obtain

$|b^{2}-7a| \ge ab^{2}.$

If $b^{2} \ge 7a$, then $b^{2}-7a \ge ab^{2}$. This implies

$b^{2} \ge ab^{2}+7a,$

hence

$b^{2}(1-a) \ge 7a,$

which is impossible for positive integers $a\ge 1$, since the left-hand side is nonpositive while the right-hand side is positive. Therefore $b^{2} \ge 7a$ cannot occur.

Hence $7a>b^{2}$, so

$7a-b^{2} \ge ab^{2}.$

This implies

$7a \ge ab^{2}+b^{2} \ge ab^{2},$

so $b^{2} \le 7$ and thus $b \in {1,2}$.

Case 3: $b=2$

The equation becomes

$4a+9 \mid 2a^{2}+a+2.$

Compute

$2(2a^{2}+a+2)-a(4a+9)=4-7a.$

Thus $4a+9 \mid 7a-4$. Another linear combination gives

$7(4a+9)-4(7a-4)=79,$

so $4a+9 \mid 79$. Since $4a+9 \ge 13$, the only possibility is $4a+9=79$, giving $a=70/4$, not an integer. No solutions occur.

Case 4: $b=1$

The condition becomes

$a+8 \mid a^{2}+a+1.$

Reducing modulo $a+8$ using $a \equiv -8$ gives

$a^{2}+a+1 \equiv 64-8+1=57 \pmod{a+8}.$

Hence $a+8 \mid 57$, so $a+8 \in {1,3,19,57}$. This yields $a=11,49$ as the positive solutions.

Case 5: $b\ge 3$

From $D \mid (b^{2}-7a)$ we have $D \le |b^{2}-7a|$. Since $7a>b^{2}$,

$ab^{2}+b+7 \le 7a-b^{2} \le 7a.$

Thus

$ab^{2} \le 7a,$

$b^{2} \le 7.$

This contradicts $b\ge 3$, so no solutions exist for $b\ge 3$.

All cases are exhausted, so the full solution set is

$\boxed{(a,b)=(7m^{2},7m)\ (m\ge 1),\ (11,1),\ (49,1)}.$

Verification of Key Steps

The identity $bN-aD=b^{2}-7a$ follows by direct expansion, with cancellation of $a^{2}b^{2}$ and $ab$ terms.

In the case $t=0$, substitution $b=7m$, $a=7m^{2}$ gives $N=mD$, establishing divisibility for all $m$.

In the case $t\ne 0$, the inequality $D \le |b^{2}-7a|$ is valid since $|t|\ge 1$. The deduction that $b^{2}\le 7$ in the branch $7a>b^{2}$ is justified by comparing growth rates, yielding a finite reduction to $b=1,2$.

For $b=2$, the elimination producing $4a+9 \mid 79$ is correct and forces non-integrality. For $b=1$, modular reduction modulo $a+8$ is valid and yields exactly two integer solutions.

Each case is fully covered, and no additional possibilities remain.

Alternative Approaches

One alternative is to parameterize directly from

$a(7+tb^{2})=b^{2}-tb-7t,$

and analyze integer constraints on $t$, showing that only $t=0$ and finitely many $t$ corresponding to $b=1$ can occur. Another approach is to bound $D$ against $|b^{2}-7a|$ to reduce the problem to finitely many small $b$ values followed by direct checking.