Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Baltic Way

Find integers $0 < a_1 < a_2 < a_3 < a_4$ such that for any $1 \le k < l \le 4$ number $a_k \cdot a_l + 1$ is a square of an integer.

Solution

Answer: For example $2$ , $4$ , $12$ , $420$ .

For $a_1 = 2$ and $a_2 = 4$ we look for a number $a_3$ such that $2a_3 + 1$ and $4a_3 + 1$ are squares, say $b^2$ and $c^2$ respectively. Then we have $2b^2 - c^2 = 1$ , which is Pell's equation.

Consider two consecutive solutions of this Pell's equation: $(5, 7)$ and $(29, 41)$ . Take $2a_3 + 1 = 5^2$ and $2a_4 + 1 = 29^2$ , which gives us $a_3 = 12$ , $a_4 = 420$ .

Now it's enough to observe that $a_3 \cdot a_4 + 1$ is a square, which is easy to count. What is more (and unnecessary in this problem) we could have taken any two consecutive solutions of this Pell's equation.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.