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Find all integers $n\geq 3$ for which there exist real numbers $a_{1}, a_{2},...,a_{n+2}$ satisfying $a_{n+1}=a_{1}$ , $a_{n+2}=a_{2}$ and:

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

for $i=1,2,..,n$

power_of_epi
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  • These olympiad problems can be really hard, but what have you tried? – Aphelli Jun 30 '19 at 20:31
  • There is a similar sequence in this question, the trick that is used there doesn't seems to works here but maybe could bring some inspiration about how to proceed. – Dabed Jun 30 '19 at 21:45

1 Answers1

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Hint.

Solving backwards we have for

$$ \begin{cases} n=2\to \{2,-1,-1,2\}\\ n=5\to \{2,-1,-1,2,\frac 12,-1,2\}\\ n=8\to \{2,-1,-1,2,\frac 12,-1,2,\frac 12,-1,2\}\\ \vdots\\ n=2+3k\to \{2,-1,\underbrace{-1,2,\frac 12}_{k},-1,2\} \end{cases} $$

NOTE

The real solutions arise from

$$ \left\{ \begin{array}{rcl} a_1&=&\frac{a_2-1}{a_1} \\ a_2&=&1-\frac{a_1}{a_2-1} \\ \end{array} \right. $$

the other possibilities

$$ \left\{ \begin{array}{c} a_1=1-\frac{a_1}{a_2-1} \\ a_2=\frac{a_1}{a_1-a_2+1} \\ \end{array} \right. \ \ \text{and}\ \ \left\{ \begin{array}{c} a_1=\frac{a_1}{a_1-a_2+1} \\ a_2=\frac{a_2-1}{a_1} \\ \end{array} \right. $$

have no real solutions.

Cesareo
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