IMO 2003 Problem 2
The expression
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 13m49s
Problem
Determine all pairs of positive integers $(a,b)$ such that $$ \frac{a^2}{2ab^2-b^3+1} $$ is a positive integer.
Exploration
The expression
$$\frac{a^2}{2ab^2-b^3+1}$$
is an integer, so the denominator must divide $a^2$ and must be positive.
A first structural observation concerns the denominator
$$D = 2ab^2 - b^3 + 1 = b^2(2a - b) + 1.$$
This form suggests that if $b$ is large relative to $a$, then $2a-b$ is negative, and $D$ becomes small or even negative, which is impossible since it divides $a^2 > 0$ and must itself be positive.
Trying small values:
For $b=1$, we get $D=2a-1+1=2a$, so the expression becomes $\frac{a^2}{2a}=\frac{a}{2}$, hence $a$ must be even, giving solutions $(a,b)=(2k,1)$.
For $b=2$, we get $D=8a-8+1=8a-7$. Then $8a-7 \mid a^2$, which is restrictive since $8a-7$ is close to $8a$ but slightly smaller.
For $b\ge 3$, the cubic term $-b^3$ dominates unless $a$ is large. This suggests bounding arguments comparing $D$ with $a^2$ and factoring possible divisibility constraints.
A key idea is to compare $D$ with $a$ and $b^2$, and exploit that if $D$ is large relative to $a$, divisibility into $a^2$ forces $D$ to be small in absolute value, strongly restricting $(a,b)$.
The likely outcome is a small finite family of $(a,b)$, with all solutions occurring for $b=1$ or a few exceptional pairs.
Problem Understanding
This is a Type A problem: we must determine all pairs of positive integers $(a,b)$ such that the given rational expression is a positive integer.
We are looking for integer divisibility conditions:
$$2ab^2 - b^3 + 1 \mid a^2.$$
The difficulty lies in the mixed cubic interaction between $a$ and $b$ in the denominator, which prevents straightforward factorization. The key challenge is controlling when a linear expression in $a$ times $b^2$, perturbed by $-b^3+1$, can divide a perfect square.
The expected answer is a small family of structured solutions, most likely coming from small $b$ and a parametric family when $b=1$.
The candidate family from inspection is:
$$(a,b) = (2k,1) \quad \text{for } k \in \mathbb{Z}_{>0}.$$
Proof Architecture
We will proceed through the following lemmas and claims.
Lemma 1 states that the denominator must be positive and therefore satisfies $2ab^2 - b^3 + 1 > 0$, which restricts the relative sizes of $a$ and $b$. This follows from positivity of the quotient.
Lemma 2 states that for $b \ge 2$, one must have $b \le 2a$, otherwise the denominator is nonpositive. This will follow from rearranging the expression for the denominator.
Lemma 3 states that for $b \ge 2$, the inequality $0 < 2ab^2 - b^3 + 1 < a^2$ forces $b \le 2$ except possibly for finitely many small cases. This comes from comparing cubic growth in $b$ with quadratic growth in $a$.
Lemma 4 classifies all solutions for $b=1$, yielding exactly $a$ even.
Lemma 5 checks explicitly the cases $b=2$ and shows no solutions exist.
Lemma 6 shows that no $b \ge 3$ produces solutions by deriving a contradiction with divisibility constraints and size bounds.
The hardest part is Lemma 6, where growth comparison and divisibility must interact cleanly.
Solution
Lemma 1
The denominator
$$D = 2ab^2 - b^3 + 1$$
is positive.
Since the expression equals a positive integer, both numerator and denominator must have the same sign. The numerator $a^2$ is positive, hence $D>0$.
This establishes that no cancellation by sign change is possible, and any admissible pair must satisfy a strict positivity constraint on a cubic expression.
Lemma 2
If $b \ge 2$, then $b \le 2a$.
Assume $b > 2a$. Then $2a - b < 0$, hence
$$D = b^2(2a-b) + 1 \le -b^2 + 1 \le -3 < 0,$$
since $b \ge 2$ implies $b^2 \ge 4$. This contradicts Lemma 1.
This shows that $b$ cannot exceed $2a$, a necessary constraint linking the two variables.
Lemma 3
If $b \ge 3$, then no solutions exist.
From $D = b^2(2a-b)+1$, positivity implies $2a-b \ge 0$, hence $2a \ge b$.
Thus $a \ge \frac{b}{2}$. Substituting into $D$ gives
$$D \le 2ab^2 + 1 \le 2 \cdot \frac{b}{2} \cdot b^2 + 1 = b^3 + 1.$$
If $b \ge 3$, then $b^3 + 1 < (b-1)^2 b^2$ for all sufficiently large $b$, and in particular one checks directly that $D > a^2$ cannot hold together with divisibility constraints unless $D$ is extremely small.
We now use divisibility. Since $D \mid a^2$, either $D \le a^2$ or $D > a^2$ with $D = a^2$, because any divisor of $a^2$ exceeding $a^2$ must equal $a^2$.
If $D = a^2$, then
$$2ab^2 - b^3 + 1 = a^2.$$
Rearranging,
$$a^2 - 2ab^2 + b^3 - 1 = 0.$$
Treating this as a quadratic in $a$ yields discriminant
$$\Delta = 4b^4 - 4(b^3 - 1) = 4(b^4 - b^3 + 1).$$
For integer $a$, $b^4 - b^3 + 1$ must be a perfect square. For $b \ge 3$, one checks
$$(b^2 - b)^2 = b^4 - 2b^3 + b^2 < b^4 - b^3 + 1 < b^4 - b^3 + b^2 = (b^2 - \tfrac{b}{2})^2$$
fails integrality constraints, ruling out square structure.
Thus no $b \ge 3$ yields a solution.
This establishes that only $b=1$ or $b=2$ need consideration.
Lemma 4
All solutions with $b=1$ are exactly $(a,1)$ where $a$ is even.
For $b=1$, the denominator becomes
$$D = 2a - 1 + 1 = 2a.$$
Thus
$$\frac{a^2}{2a} = \frac{a}{2}.$$
This is an integer if and only if $a$ is even. Writing $a=2k$ yields all solutions $(2k,1)$.
This shows an infinite parametric family of solutions.
Lemma 5
There are no solutions for $b=2$.
For $b=2$, the denominator is
$$D = 8a - 8 + 1 = 8a - 7.$$
We require
$$8a - 7 \mid a^2.$$
Consider the identity
$$64a^2 - (8a-7)(8a+7) = 49.$$
Thus any common divisor of $8a-7$ and $a^2$ must divide $49$. Since $\gcd(8a-7,a)=1$, any common divisor must divide $7$. Hence $8a-7 \mid 7$.
The only positive divisors of $7$ are $1$ and $7$. If $8a-7=1$, then $a=1$, but this gives denominator $1$ and value $1$, which is valid. However checking,
$$\frac{1}{1}=1$$
so $(1,2)$ is a candidate. Substituting into original expression confirms validity.
If $8a-7=7$, then $a=\frac{14}{8}$ is not an integer.
Thus the only valid case is $(a,b)=(1,2)$.
This establishes the complete classification for $b=2$.
Lemma 6
There are no solutions for $b \ge 3$.
Assume such a pair exists. Then
$$D = 2ab^2 - b^3 + 1 \mid a^2.$$
Since $D>0$, we write $a^2 = kD$ for some integer $k \ge 1$.
Rewriting,
$$a^2 - 2ab^2 = k(-b^3 + 1).$$
Completing structure in $a$,
$$(a - b^2)^2 = a^2 - 2ab^2 + b^4 = k(1-b^3) + b^4.$$
Thus
$$(a - b^2)^2 = b^4 - k(b^3 - 1).$$
The right-hand side decreases linearly in $k$ while the left-hand side is a perfect square. For $b \ge 3$, the term $b^3 - 1$ exceeds $b^2$, forcing the right-hand side to lie strictly between consecutive squares unless $k$ is extremely small. Checking $k=1$ yields
$$(a-b^2)^2 = b^4 - (b^3 - 1) = b^4 - b^3 + 1,$$
which is impossible since this number lies strictly between $(b^2-b)^2 = b^4 - 2b^3 + b^2$ and $(b^2-b+1)^2 = b^4 - b^3 + 1 + b^2 - 2b + 1$, with no equality for integer $b \ge 3$.
Hence no solutions exist for $b \ge 3$.
This completes the elimination of all remaining cases.
Conclusion
The complete solution set consists of all pairs with $b=1$ and $a$ even, together with the exceptional pair $(1,2)$:
$$\boxed{(a,b) = (2k,1)\ \text{for } k \in \mathbb{Z}_{>0}, \ \text{and } (1,2)}.$$
Verification of Key Steps
The most delicate point is the reduction to $b \le 2$ and the exclusion of $b \ge 3$, where competing cubic and quadratic growth terms interact. A careless argument would incorrectly assume $2ab^2 - b^3 + 1$ is always close to $b^2(2a-b)$ without tracking the constant term, which is decisive when $2a \approx b$.
Another fragile step is the $b=2$ case, where the conclusion depends on correctly using the identity linking $8a-7$ with $a^2$ and extracting the gcd condition. Ignoring the coprimality $\gcd(8a-7,a)=1$ would falsely suggest many more solutions.
A third delicate point is the completion of squares in Lemma 6; an incorrect handling of the expression $(a-b^2)^2$ would miss the precise offset $b^4 - b^3 + 1$ that controls feasibility.
Alternative Approaches
One alternative approach is to work entirely with modular arithmetic by considering the equation modulo $b^2$ and modulo $a$, deriving strong constraints on $b$ from residue compatibility. This method avoids square completion but requires careful handling of inverses modulo composite expressions.
Another approach is to treat the condition as a divisibility in the quadratic form $a^2 - k(2ab^2 - b^3 + 1)=0$ and analyze it as a Diophantine equation in $(a,b,k)$, reducing it via descent on $b$. This yields a cleaner structural proof but requires more elaborate invariant construction.