HMMT, 2002

Algebra Difficulty 5.0 AIME Prove it United States

Problem:

For positive integers $L$ , let $S_{L} = \sum_{n=1}^{L} \lfloor n / 2 \rfloor$ . Determine all $L$ for which $S_{L}$ is a square number.

Solution

Solution:

We distinguish two cases depending on the parity of $L$ .

Suppose first that $L = 2k - 1$ is odd, where $k \geq 1$ . Then
$S_{L} = \sum_{1 \leq n \leq 2k-1} \left\lfloor \frac{n}{2} \right\rfloor = 2 \sum_{0 \leq m < k} m = 2 \cdot \frac{k(k-1)}{2} = k(k-1).$
If $k = 1$ , this is the square number $0$ . If $k > 1$ then $(k-1)^2 < k(k-1) < k^2$ , so $k(k-1)$ is not square.

Now suppose $L = 2k$ is even, where $k \geq 1$ . Then $S_{L} = S_{L-1} + k = k^2$ is always square.

Hence $S_{L}$ is square exactly when $L = 1$ or $L$ is even.

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