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 4 of 7: A vanishing sum of squares forces period

3

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\sum_{i=1}^n (a_i-a_{i+3})^2=0\ \Longrightarrow\ a_i=a_{i+3}\ (1\le i\le n)

Detailed analysis

Since the extension is $n$ -periodic, $\sum_{i=1}^n a_{i+3}^2=\sum_{i=1}^n a_i^2$ as well. Expanding $(a_i-a_{i+3})^2=a_i^2-2a_ia_{i+3}+a_{i+3}^2$ and summing over $i=1,\ldots,n$ , the identity of Step 3 gives $\sum_{i=1}^n (a_i-a_{i+3})^2=2\sum_{i=1}^n a_i^2-2\sum_{i=1}^n a_ia_{i+3}=0$ . A finite sum of squares of real numbers vanishes only if every term vanishes, so $a_i=a_{i+3}$ for every $i=1,\ldots,n$ (indices read modulo $n$ ).