Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it China

Find a triple $(l, m, n)$ ( $1 < l < m < n$ ) of positive integers such that $\sum_{k=1}^{l} k$ , $\sum_{k=l+1}^{m} k$ , $\sum_{k=m+1}^{n} k$ form a geometric sequence in order. (posed by Tao Pingsheng)

Solution

$\sum_{k=1}^{l} k = S_{l}$ , $\sum_{k=l+1}^{m} k = S_{m} - S_{l}$ , $\sum_{k=m+1}^{n} k = S_{n} - S_{m}$ ,
form a geometric sequence in order. Then
$S_{l}(S_{n} - S_{m}) = (S_{m} - S_{l})^2, \qquad \textcircled{1}$
that is $S_{l}(S_{n} + S_{m} - S_{l}) = S_{m}^2$ . Thus, $S_{l} \mid S_{m}^2$ , that is
$2l(l + 1) \mid m^2(m + 1)^2.$
Let $m + 1 = l(l + 1)$ and take $l = 3$ . Then $m = 11$ and $S_{l} = S_{3} = 6$ , $S_{m} = S_{11} = 66$ . Substitute it into ①, we have $S_{n} = 666$ , that is $\frac{n(n+1)}{2} = 666$ . So $n = 36$ .
Therefore, $(l, m, n) = (3, 11, 36)$ is a solution satisfying the condition. □

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