2007 IMO Problems/Problem 5

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 $a=b$.

Solution

Lemma. If there is a counterexample for some value of $a$, then there is a counterexample $(a,b)$ for this value of $a$ such that $b<a$.

Proof. Suppose that $b >a$. Note that $4ab -1 \equiv -1 \pmod{4a}$, but $(4a^2-1)^2 \equiv 1 \pmod{4a}$. It follows that $(4a^2-1)^2/(4ab-1) \equiv -1 \pmod{4a}$. Since \[0<(4a^2-1)^2/(4ab-1) < (4a^2-1)^2/(4a^2-1) = 4a^2-1,\] it follows that $(4a^2-1)^2/(4ab-1)$ can be written as $4ab'-1$, with $0<b'<a$. Then $(a,b')$ is a counterexample for which $b'<a$. $\blacksquare$

Now, suppose a counterexample exists. Let $(a,b)$ be a counterexample for which $a$ is minimal and $b<a$. We note that \[\gcd(4ab-1,2a-1) \mid 4ab-1 - 2b(2a-1) = 2b-1,\] and \[\gcd(4ab-1,2a+1) \mid 2b(2a+1) - (4ab-1) = 2b+1 .\] Now, \begin{align*} 4ab-1 &\mid (4a^2-1)^2 = (2a-1)^2(2a+1)^2 \\ &\mid \gcd(4ab-1,2a-1)^2 \cdot \gcd(4ab-1,2a+1)^2 \\ &\mid (2b-1)^2(2b+1)^2 = (4b^2-1)^2 . \end{align*} Thus $(b,a)$ is a counterexample. But $b<a$, which contradicts the minimality of $a$. Therefore no counterexample exists. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

2007 IMO (Problems) • Resources)
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Problem 4
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Problem 6
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