Question

Let \(\{A_n\}\) be a unique sequence of positive integers satisfying the following properties: \(A_1 = 1,\ A_2 = 2,\ A_4 = 12\), and \(A_{n+1}\cdot A_{n-1} = A_n^2 \pm 1\) for \(n = 2, 3, 4, \dots\)
Then, \(A_7\) is:

• A. 60
• B. 120
• C. 149
• D. 169

Hint:

At each step, try both the +1 and -1 sign in A_(n+1)A_(n-1) = A_n^2 plus-or-minus 1, and keep only the choice that gives a positive integer; use the given A4 = 12 to fix A3 first.

Answer 1

The Correct Option is D

Step 1: Understand the recursion.
At each step n, the rule \(A_{n+1}A_{n-1} = A_n^2 \pm 1\) leaves a choice of sign; the problem says exactly one choice at each step keeps the whole sequence made of positive integers and matches the given values \(A_1=1,\ A_2=2,\ A_4=12\). We find the right sign at each step by testing which one gives an integer.

Step 2: Find A3 using n = 2.
\[ A_3 A_1 = A_2^2 \pm 1 = 4 \pm 1 \]
This gives \(A_3 = 5\) (using +1) or \(A_3 = 3\) (using \(-1\)), since \(A_1=1\). Both are positive integers, so we need the n=3 relation, using the known value \(A_4=12\), to decide which is correct.

Step 3: Fix A3 using n = 3 and the given A4 = 12.
\[ A_4 A_2 = A_3^2 \pm 1 \implies 12 \times 2 = 24 = A_3^2 \pm 1 \]
If \(A_3=5\): \(A_3^2=25\), and \(25-1=24\). This matches (using the \(-1\) sign).
If \(A_3=3\): \(A_3^2=9\), and \(9\pm1\) is 8 or 10, neither equal to 24. This does not work.
So \(A_3 = 5\) is the only value consistent with \(A_4=12\).

Step 4: Find A5 using n = 4.
\[ A_5 A_3 = A_4^2 \pm 1 \implies 5 A_5 = 144 \pm 1 = 145 \text{ or } 143 \]
\(145/5 = 29\) (integer), while \(143/5\) is not an integer. So \(A_5 = 29\).

Step 5: Find A6 using n = 5.
\[ A_6 A_4 = A_5^2 \pm 1 \implies 12 A_6 = 841 \pm 1 = 842 \text{ or } 840 \]
\(840/12 = 70\) (integer), while \(842/12\) is not. So \(A_6 = 70\).

Step 6: Find A7 using n = 6.
\[ A_7 A_5 = A_6^2 \pm 1 \implies 29 A_7 = 4900 \pm 1 = 4901 \text{ or } 4899 \]
\(4901 / 29 = 169\) (integer, since \(29 \times 169 = 4901\)), while \(4899/29\) is not a whole number. So \(A_7 = 169\).

Final Answer:
\[ \boxed{A_7 = 169} \]