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 3 of 7: Summing the identity cyclically

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\sum_{i=1}^n a_i^2=\sum_{i=1}^n a_ia_{i+3}

Detailed analysis

Sum the identity from Step 2 over $i=1,\ldots,n$ . Because the extension is $n$ -periodic, the index $i+2$ runs through the same $n$ values as $i$ does, so $\sum_{i=1}^n a_{i+2}^2=\sum_{i=1}^n a_i^2$ and $\sum_{i=1}^n a_{i+2}=\sum_{i=1}^n a_i$ ; hence the linear terms $-a_{i+2}$ on the left and $-a_i$ on the right of the summed identity cancel, leaving $\sum_{i=1}^n a_i^2=\sum_{i=1}^n a_ia_{i+3}$ .