Competition problem — Number theory

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Number theory Difficulty 5.6 AIME, harder Prove it

Let $A$ and $B$ be positive integers. Define the arithmetic sequence $a_{0}, a_{1}, a_{2}, \ldots$ by $a_{n}=A n+B$ . Assume that there is at least one $n \geq 0$ such that $a_{n}$ is a square. Let $M$ be a positive integer such that $M^{2}$ is the smallest square in the sequence. Prove that $M<A+\sqrt{B}$ .

Solution

If $M \leq A$ , then certainly $MA$ , let there be $k$ such that $a_{k}=M^{2}$ . Then it holds that $A k+B=M^{2}$ . Since $0<M-A<$ $M$ , $(M-A)^{2}$ is smaller than $M^{2}$ . It holds that $(M-A)^{2}=M^{2}-2 M A+A^{2}=M^{2}-A(2 M-A)$ . If $k-(2 M-A) \geq 0$ , then
$a_{k-(2 M-A)}=A(k-(2 M-A))+B=(A k+B)-2 M A+A^{2}=M^{2}-2 M A+A^{2}=(M-A)^{2}$ .
But then $M^{2}$ is not the smallest square in the sequence, contradiction. Therefore, $k-(2 M-A)<$ 0. This implies $A(k-(2 M-A))+B<B$ , or $(M-A)^{2}<B$ . Since $M-A$ is positive, we can conclude that $M-A<\sqrt{B}$ , or $M<A+\sqrt{B}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.