Mid 21st century

Algebra Level 4

For which of the following values of $n$ does there exist real numbers $a_1, a_2, . . . , a_n, a_{n+1} = a_1, a_{n+2} = a_2$ such that

$\large\ { a }_{ i }{ a }_{ i+1 } + 1 = { a }_{ i+2 }$

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

2019 2017 2018 2021 2020

1 solution

Mark Hennings
Aug 2, 2019

The sequence $a_j \; = \; \left\{ \begin{array}{lll} -1 & \hspace{1cm} & j \equiv 0,1 \pmod{3} \\ 2 & & j \equiv 2 \pmod{3} \end{array} \right.$ is a sequence of period $3$ which satisfies the condition $a_{j+2} \; = \; a_{j+1}a_j + 1 \hspace{2cm} j \ge 1$ The sequence is therefore of period $3m$ for any integer $m$ . Thus we can certainly find an acceptable sequence of period $n$ provided that $3$ divides $n$ . Of the options given, $\boxed{2019}$ is the only multiple of $3$ .

Sir, how did you arrive at the sequence in the first place??

Aaghaz Mahajan - 1 year, 10 months ago

To be honest... trial and error! No periodic sequence can have two successive positive terms, and I played around for a while.

Mark Hennings - 1 year, 10 months ago

Ohhhh!!! That's good to know!!! Because, I solved this the same way and was wondering whether I got lucky or there is some logical explanation.....:P

Aaghaz Mahajan - 1 year, 10 months ago

@Aaghaz Mahajan – Well, we may have solved the question, but we have not really solved the problem. We need to show that there is not another solution of a period coprime to $3$ ...

Mark Hennings - 1 year, 10 months ago

Mid 21st century

1 solution

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