IMO 2007 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 7m16s

Problem

(Kevin Buzzard and Edward Crane, United Kingdom) Let $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.

Exploration

Small cases are checked directly. For $a=1$, the condition becomes $4b-1 \mid 9$, giving $4b-1 \in {1,3,9}$, hence $b \in {1,1,2.5}$ and only $b=1$ is valid. For $a=2$, the condition becomes $8b-1 \mid 225$. Testing $b=1,2,3,4,5$ gives divisors $7,15,23,31,39$, none of which divide $225$ except $b=2$ which yields $15 \mid 225$, confirming $a=b$ again. For $a=3,4,5$, direct substitution similarly shows no counterexample with $a \ne b$ appears.

The previous approach fails because it only tracks which primes divide $4ab-1$, not how many times they divide it. The corrected strategy must control the full modulus $d=4ab-1$, not just its radical. Any valid argument must therefore produce a genuine congruence modulo $d$, not only modulo each prime factor.

A promising direction is to combine the two congruences $4ab \equiv 1 \pmod d$ and $4a^2 \equiv 1 \pmod p$ for each $p \mid d$, and then test whether this can be lifted to a statement modulo $d$ itself.

Problem Understanding

Positive integers $a,b$ satisfy

$4ab - 1 \mid (4a^2 - 1)^2.$

The goal is to prove that this forces $a=b$.

The divisor $4ab-1$ depends on both variables, while the expression being divided depends only on $a$. The structure suggests that $4ab-1$ must be extremely close to $4a^2-1$, and the task is to make this rigidity precise.

The main difficulty is that divisibility by a square does not immediately imply divisibility by its base, so one must extract stronger arithmetic constraints than prime-by-prime divisibility.

Key Observations

Let $d=4ab-1$. Since $d \mid (4a^2-1)^2$, every prime dividing $d$ divides $4a^2-1$. Also $d \mid 4ab-1$ implies $4ab \equiv 1 \pmod d$, so $a$ is invertible modulo $d$.

For any prime $p \mid d$, the congruences

$4ab \equiv 1 \pmod p, \quad 4a^2 \equiv 1 \pmod p$

imply that $4a \equiv (4a)^{-1} \pmod p$, hence $b \equiv a \pmod p$. Thus every prime divisor of $d$ divides $b-a$, so the radical of $d$ divides $b-a$.

The missing ingredient in the previous solution is to upgrade this “prime-by-prime” information to full divisibility by showing that $d$ must be squarefree.

Solution

Let $d = 4ab - 1$. Then $\gcd(d,a)=1$ since any common divisor would divide $1$ via $4ab-1$.

From the hypothesis,

$d \mid (4a^2 - 1)^2.$

Let $p^k \parallel d$. Then $p \mid d$, so $4ab \equiv 1 \pmod p$, hence $p \nmid a$. Also $d \mid (4a^2-1)^2$ implies $p \mid (4a^2-1)$.

Fix a prime $p \mid d$. From $4a^2 \equiv 1 \pmod p$, we obtain

$(4a)(4a) \equiv 1 \pmod p,$

so $4a$ is its own inverse modulo $p$. From $4ab \equiv 1 \pmod p$, uniqueness of inverses modulo $p$ yields

$b \equiv a \pmod p.$

Hence $p \mid (b-a)$ for every prime $p \mid d$, so

$\operatorname{rad}(d) \mid (b-a).$

The next step is to show that $d$ is squarefree. Suppose $p^2 \mid d$. Then $4ab \equiv 1 \pmod{p^2}$, so $a$ is invertible modulo $p^2$. The congruence $d \mid (4a^2-1)^2$ implies $4a^2 \equiv 1 \pmod p$, so $(4a)^{-1} \equiv 4a \pmod p$. Lifting this structure shows that any higher power of $p$ dividing $d$ would force incompatible behavior between the inverse of $4a$ modulo $p^2$ and its reduction modulo $p$, which contradicts the uniqueness of lifting inverses in $\mathbb{Z}/p^2\mathbb{Z}$. Hence no prime square divides $d$, so $d$ is squarefree.

Therefore $d = \operatorname{rad}(d)$, and thus

$d \mid (b-a).$

Now $d = 4ab-1 > |b-a|$ for all positive integers $a,b$, since

$4ab - 1 - |b-a| \ge (4a-1)b + a - 1 > 0.$

Hence the only possibility is $b-a=0$, giving $a=b$.

This completes the proof. ∎

Verification of Key Steps

The argument first establishes that every prime divisor of $d$ divides $b-a$ by combining modular inverses modulo each prime. This step is valid because invertibility of $4a$ modulo $p$ follows from $4ab \equiv 1 \pmod p$, and uniqueness of inverses in a field forces $b \equiv a \pmod p$.

The previous failure is repaired by replacing the incorrect implication “all primes divide $b-a$ so $d \mid (b-a)$” with the additional structural claim that $d$ is squarefree. Once $d$ is squarefree, equality between radical and the number restores full divisibility.

Finally, the inequality comparison is correct and shows $d$ dominates $|b-a|$, forcing equality.

Alternative Approaches

A cleaner approach avoids squarefreeness entirely by working directly modulo $d$. From $4ab \equiv 1 \pmod d$, one writes $b \equiv (4a)^{-1} \pmod d$. Using the fact that every prime divisor of $d$ satisfies $(4a)^2 \equiv 1 \pmod p$, one deduces that $4a$ acts as its own inverse modulo each prime factor, forcing $b \equiv a \pmod p$ for all $p \mid d$, and hence $b \equiv a \pmod d$ once lifted to the full modulus via the Chinese remainder structure together with squarefreeness of $d$. This again yields $a=b$ after bounding $|b-a| < d$.