Problem 2168

Let $S_{n-2} = \sum_{k=1}^{n-2} \frac{1}{a_k a_{k+2}}$ and $Q_m = \sum_{k=1}^m a_k^2$. The given relation is $S_{n-2} = 1 - \frac{1}{Q_{n-1}}$ for $n \ge 3$.

For $n = 3$:
$$\frac{1}{a_1 a_3} = 1 - \frac{1}{a_1^2 + a_2^2} \quad (1)$$
This implies $a_1 = 1$. If $a_1 \ge 2$, then since $a_i \in \mathbb{N}$, $a_3 \ge 1$. So $\frac{1}{a_1 a_3} \le \frac{1}{2 \cdot 1} = \frac{1}{2}$. Also, $a_1^2 + a_2^2 \ge 2^2 + 1^2 = 5$. So $1 - \frac{1}{a_1^2 + a_2^2} \ge 1 - \frac{1}{5} = \frac{4}{5}$. Thus $\frac{1}{2} \ge \frac{4}{5}$, which is a contradiction. Yielding $a_1 = 1$.

Moreover, $a_3 \ne 1$. If $a_3 = 1$ (and $a_1 = 1$), then $\frac{1}{1 \cdot 1} = 1$. The equation becomes $1 = 1 - \frac{1}{1^2+a_2^2}$, which implies $\frac{1}{1+a_2^2} = 0$, impossible.

Now we show $a_2 = 1$. Assume $a_2 \ge 2$ (we know $a_1 = 1$). LHS of (1) is $\frac{1}{a_1 a_3} = \frac{1}{a_3}$. Since $a_3 \ge 2$ (as $a_3 \ne 1$), $\frac{1}{a_3} \le \frac{1}{2}$. RHS of (1) is $1 - \frac{1}{a_1^2+a_2^2} = 1 - \frac{1}{1^2+a_2^2}$. If $a_2 \ge 2$, then $1+a_2^2 \ge 1+2^2 = 5$. So $1 - \frac{1}{1+a_2^2} \ge 1 - \frac{1}{5} = \frac{4}{5}$. Thus $\frac{1}{2} \ge \frac{4}{5}$, a contradiction. So $a_2 = 1$.

Substituting $a_1 = 1, a_2 = 1$ into (1): $\frac{1}{a_3} = 1 - \frac{1}{1^2+1^2} = 1 - \frac{1}{2} = \frac{1}{2}$. So $a_3 = 2$.

For $n = 4$: The relation is $\frac{1}{a_1 a_3} + \frac{1}{a_2 a_4} = 1 - \frac{1}{a_1^2 + a_2^2 + a_3^2}$. Using $a_1 = 1, a_2 = 1, a_3 = 2$: $\frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 4} = 1 - \frac{1}{1^2+1^2+2^2} = 1 - \frac{1}{1+1+4} = 1 - \frac{1}{6} = \frac{5}{6}$. $\frac{1}{2} + \frac{1}{4} = \frac{5}{6} \implies \frac{1}{4} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$. So $a_4 = 3$.

For $n = 5$, we have $\frac{1}{a_1 a_3} + \frac{1}{a_2 a_4} + \frac{1}{a_3 a_5} = 1 - \frac{1}{a_1^2 + a_2^2 + a_3^2 + a_4^2}$. The sum of the first two terms is $\frac{5}{6}$ (from $n = 4$ calculation). $\frac{5}{6} + \frac{1}{a_3 a_5} = 1 - \frac{1}{1^2+1^2+2^2+3^2} = 1 - \frac{1}{1+1+4+9} = 1 - \frac{1}{15} = \frac{14}{15}$. $\frac{1}{a_3 a_5} = \frac{14}{15} - \frac{5}{6} = \frac{28-25}{30} = \frac{3}{30} = \frac{1}{10}$. Since $a_3 = 2$, $\frac{1}{2a_5} = \frac{1}{10} \implies 2a_5 = 10 \implies a_5 = 5$.

The sequence starts 1, 1, 2, 3, 5, ... This appears to be the Fibonacci sequence, where $a_n = F_n$ (with $F_1 = 1, F_2 = 1$). We claim that $a_n = a_{n-1} + a_{n-2}$ for $n \ge 3$. The right side values are $1 - 1/Q_2 = 1/2$ (for $n = 3$), $1 - 1/Q_3 = 5/6$ (for $n = 4$), $1 - 1/Q_4 = 14/15$ (for $n = 5$). This suggests the identity $Q_m = \sum_{k=1}^m a_k^2 = a_m a_{m+1}$ for the Fibonacci sequence. So the given relation would be $\sum_{k=1}^{n-2} \frac{1}{a_k a_{k+2}} = 1 - \frac{1}{a_{n-1} a_n}$.

We prove this claim by induction. The base case $n = 3$ holds: $\frac{1}{a_1 a_3} = \frac{1}{1 \cdot 2} = \frac{1}{2}$. And $1 - \frac{1}{a_2 a_3} = 1 - \frac{1}{1 \cdot 2} = \frac{1}{2}$. Assume the relation holds for $n-1$, meaning $\sum_{k=1}^{n-3} \frac{1}{a_k a_{k+2}} = 1 - \frac{1}{a_{n-2} a_{n-1}}$, and assume $a_k$ are Fibonacci numbers up to $n-1$, and $Q_m = a_m a_{m+1}$ for $m \le n-2$. We define $a_n = a_{n-1} + a_{n-2}$. We need to show the statement holds for $n$. LHS for $n$: $\sum_{k=1}^{n-2} \frac{1}{a_k a_{k+2}} = (\sum_{k=1}^{n-3} \frac{1}{a_k a_{k+2}}) + \frac{1}{a_{n-2} a_n}$. Using the inductive hypothesis for the sum: LHS = $(1 - \frac{1}{a_{n-2} a_{n-1}}) + \frac{1}{a_{n-2} a_n} = 1 - \frac{1}{a_{n-2}} (\frac{1}{a_{n-1}} - \frac{1}{a_n})$

LHS = $1 - \frac{1}{a_{n-2}} (\frac{a_n - a_{n-1}}{a_{n-1} a_n})$. Since $a_n = a_{n-1} + a_{n-2}$, we have $a_n - a_{n-1} = a_{n-2}$. LHS = $1 - \frac{1}{a_{n-2}} (\frac{a_{n-2}}{a_{n-1} a_n}) = 1 - \frac{1}{a_{n-1} a_n}$.

For the RHS, we need to verify $Q_{n-1} = a_{n-1} a_n$. $Q_{n-1} = \sum_{k=1}^{n-1} a_k^2 = (\sum_{k=1}^{n-2} a_k^2) + a_{n-1}^2 = Q_{n-2} + a_{n-1}^2$. By the inductive hypothesis on the sum of squares formula, $Q_{n-2} = a_{n-2} a_{n-1}$. So $Q_{n-1} = a_{n-2} a_{n-1} + a_{n-1}^2 = a_{n-1}(a_{n-2} + a_{n-1})$. Since $a_n = a_{n-1} + a_{n-2}$, $Q_{n-1} = a_{n-1} a_n$. So the RHS is $1 - \frac{1}{Q_{n-1}} = 1 - \frac{1}{a_{n-1} a_n}$. The relation holds for $n$ if we define $a_n$ using the Fibonacci recurrence. The initial terms $a_1 = 1, a_2 = 1$ were uniquely determined. The subsequent terms $a_3, a_4, a_5$ were also uniquely determined by the relation. The Fibonacci sequence $F_n$ (with $F_1 = 1, F_2 = 1$) matches these terms and satisfies the relation for all $n \ge 3$. Thus, the only solution is the Fibonacci sequence $a_n = F_n (1, 1, 2, 3, 5, \dots)$. ■