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Berkeley Math Circle Monthly Contest 4

Algebra Difficulty 4.9 AIME Prove it United States

Problem:

A natural number $n$ is chosen between two consecutive square numbers. The smaller square is obtained by subtracting $k$ from $n$ , and the larger one is obtained by adding $\ell$ to $n$ . Prove that the number $n-k \ell$ is the square of an integer.

Solution

Solution:

Let $n-k = x^{2}$ and $n+\ell = (x+1)^{2}$ . Then $k = n - x^{2}$ and $\ell = (x+1)^{2} - n$ . We express everything in terms of $n$ and $x$ :
$\begin{aligned} n - k \ell & = n - \left(n - x^{2}\right)\left((x+1)^{2} - n\right) \\ & = n + \left(x^{2} - n\right)\left(x^{2} + 2x + 1 - n\right) \\ & = n + \left(x^{2} - n\right)\left(\left(x^{2} - n\right) + 2x + 1\right) \\ & = n + \left(x^{2} - n\right)^{2} + 2x\left(x^{2} - n\right) + x^{2} - n \\ & = \left(x^{2} - n\right)^{2} + 2x\left(x^{2} - n\right) + x^{2} \\ & = \left(x^{2} - n + x\right)^{2}. \end{aligned}$
This is the square of an integer.

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