2022 USAMO Problems/Problem 4

Problem

Solution

Since $q(p-1)$ is a perfect square and $q$ is prime, we should have $p - 1 = qb^2$ for some positive integer $b$ . Let $a^2 = p - q$ . Therefore, $q = p - a^2$ , and substituting that into the $p - 1 = qb^2$ and solving for $p$ gives $p = \frac{a^2b^2 - 1}{b^2 - 1} = \frac{(ab - 1)(ab + 1)}{b^2 - 1}.$ Notice that we also have $p = \frac{a^2b^2 - 1}{b^2 - 1} = a^2 + \frac{a^2 - 1}{b^2 - 1}$ and so $b^2 - 1 | a^2 - 1$ . We run through the cases

$a = 1$ : Then $p - q = 1$ so $(p, q) = (3, 2)$ , which works.
$a = b$ : This means $p = a^2 + 1$ , so $q = 1$ , a contradiction.
$a > b$ : This means that $b^2 - 1 < ab - 1$ . Since $b^2 - 1$ can be split up into two factors $F_1, F_2$ such that $F_1 | ab - 1$ and $F_2 | ab + 1$ , we get

$p = \frac{ab - 1}{F_1} \cdot \frac{ab + 1}{F_2}$ and each factor is greater than $1$ , contradicting the primality of $p$ .

Thus, the only solution is $(p, q) = (3, 2)$ .

2022 USAMO (Problems • Resources)
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2022 USAMO Problems/Problem 4

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