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. $\blacksquare$

(Sean Yu, US)

Second Solution

As $(4a^2-1)^2 = (4a^2-4ab + 4ab-1)^2 \equiv (4a)^2(a-b)^2 \pmod {4ab-1}$ , since $gcd((4a)^2, 4ab -1 ) = 1$ , we only need to show that $4ab -1 \mid (a-b)^2 \Rightarrow a=b$ . We will use the Vieta Jumping technique to solve it.

Suppose there exist counterexamples, which means the set $S = \{(x, y): \mbox {x and y are positive integers}, x \ne y, 4xy-1 \mid (x-y)^2\}$ is not empty. Then there exists a pair $(a, b)\in S$ such that $a+b$ has the minimum sum among all pairs in $S$ . Without loss of generality, we assume $a > b$ . Let $k=\frac{(a-b)^2}{4ab-1}$ . Then $k$ is a positive integer and $a^2-(2+4k)ba + b^2+k = 0$ . So $a$ is a root for the quadratic equation $x^2-(2+4k)bx + b^2+k = 0$ . Using the Vieta Lemma, we know the equation has another root $a' = \frac{b^2 + k}{a} > 0$ and from $a' = (2+4k)b -a$ we know it is an integer. Note that since $4ab -1 \mid(a-b)^2$ and $a>b$ , we have $4ab -1 \le (a-b)^2 \le a^2-1$ , which implies $b\le a/4$ . Also $k=\frac{(a-b)^2}{4ab-1} \le \frac{a^2}{3ab} \le a/3$ , thus $a' = \frac{b^2 + k}{a} \le \frac{(a/4)^2 + a/3}{a} = a/16 + 1/3 < a$ , and we get another pair $(a', b)\in S$ , but $a' + b < a+b$ , which contradicts assumption that $a+b$ has the minimum sum. $\hfill\blacksquare$

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

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2007 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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2007 IMO Problems/Problem 5

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Problem

Solution

Second Solution

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