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#1 h Apr 27, 2025, 1:03 PM

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Let $k$ be a positive integer. Determine all sequences $(a_n)_{n\geq 1}$ of positive integers such that
$a_{n+2}(a_{n+1} - k) = a_n(a_{n+1} + k)$ for all positive integers $n$ .

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internationalnick123456

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internationalnick123456

#2 h Apr 30, 2025, 4:04 PM

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First, observe that $a_n > k,\forall n\in\mathbb N$ .
If there exists some $m$ such that $a_{m+1} = a_m + k$ , then by induction it follows that $a_{n+1} = a_n + k ,\forall n\in\mathbb N$ , which forms a valid sequence.
Now assume for the sake of contradiction that $a_{n+1} \neq a_n + k,\forall n \in \mathbb{N}$ . Then, we have:
$\frac{a_{n+2}}{a_{n+1} + k} = \frac{a_n}{a_{n+1} - k}\Leftrightarrow \frac{a_{n+2}}{a_{n+2} - a_{n+1} - k} = \frac{a_n}{a_n + k - a_{n+1}}$ $\Leftrightarrow \frac{a_{n+2} - a_{n+1} - k}{a_{n+1} - a_n - k} = -\frac{a_{n+2}}{a_n}$ Hence, we obtain
$\frac{a_{n+2} - a_{n+1} - k}{a_2 - a_1 - k} = (-1)^n \cdot \frac{a_{n+2} a_{n+1}}{a_2 a_1},\forall n\in\mathbb N$ $\Rightarrow |a_{n+2} - a_{n+1} - k| = c \cdot a_{n+2} a_{n+1}$ , where $c = \left| \dfrac{a_2 - a_1 - k}{a_2 a_1} \right|$
However, since $a_{n+2} > a_n$ , it follows that $\displaystyle \lim_{n \to \infty} a_n = +\infty$ . For sufficiently large $n$ , the left-hand side grows linearly while the right-hand side grows quadratically:
$|a_{n+2} - a_{n+1} - k| < 2 \max\{a_{n+2}, a_{n+1}\} < c \cdot a_{n+2} a_{n+1}$ which is a contradiction.
Thus, $a_{n+1} = a_n + k,\forall n\in\mathbb N$ .

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