Problem 2

Find all integers

n \ge 3

for which there exist real numbers

a_1, a_2, \ldots, a_n

satisfying

a_{n+1}=a_1

a_{n+2}=a_2

, and

a_ia_{i+1}+1=a_{i+2}

for

i=1,2,\ldots,n

Step 2 of 7: Extending the indices and a product identity

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a_{i+2}^2-a_{i+2}=a_ia_{i+3}-a_i

Detailed analysis

Extend $(a_i)$ to all integers $i$ by $a_{i+n}=a_i$ , consistent with $a_{n+1}=a_1$ and $a_{n+2}=a_2$ ; then $a_ia_{i+1}+1=a_{i+2}$ holds for every integer $i$ . Multiplying this relation by $a_{i+2}$ gives $a_ia_{i+1}a_{i+2}=a_{i+2}^2-a_{i+2}$ ; multiplying the shifted relation $a_{i+1}a_{i+2}+1=a_{i+3}$ by $a_i$ gives the same product, $a_ia_{i+1}a_{i+2}=a_ia_{i+3}-a_i$ . Equating the two expressions yields $a_{i+2}^2-a_{i+2}=a_ia_{i+3}-a_i$ for every integer $i$ .