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Question Number 74614 by chess1 last updated on 27/Nov/19

	Answered by mind is power last updated on 27/Nov/19
	$we can extend without and U_(i+n) =a_i ,∀i∈[1,n] U_i .U_(i+1) +1=U_(i+2) ⇒a_(n+3) =a_3 ,if a_1 ,.......a_n the sequence U_n exist since is (a_(1,) ,a_2 ,.....a_n )is repeted infintly times if U_n exist ⇒a_i existe tack just n+2 first therm so we can extend a_i without lost a_(i+1) a_i +1=a_(i+2) { ((a_(i+2) a_(i+1) +1=a_(i+3) ......1)),((a_i a_(i+1) +1=a_(i+2) ...2)) :} 1∗a_i −..2∗a_(i+2) ⇒a_i a_(i+3) −a_(i+2) ^2 =a_i −a_(i+2) Σ_(i=1) ^n (a_i a_(i+3) −a_(i+2) ^2 )=Σ_(i=1) ^n a_i −a_(i+2) =a_1 +a_2 −a_(n+2) −a_(n+1) =0 ⇒Σ_(i=1) ^n (a_i a_(i+3) −a_(i+2) ^2 )=0...E Σ_(i=1) ^n a_(i+2) =Σ_(k=1) ^n a_k ^2 ,since a_(n+1) =a_1 &a_(n+2) =a_2 ⇒E⇔Σ_(k=1) ^n (a_k a_(k+3) −a_k ^2 )=0 Σ_(k=1) ^n a_(k+3) ^2 =Σ_(k=1) ^n a_k ^2 −a_3 ^2 +a_(n+3) ^2 ,withe our expended series =Σ_(k=1) ^n a_k ^2 ⇒E⇔2Σ_(k=1) ^n a_k a_(k+3) −2Σ_(k=1) ^n a_k ^2 =0 ⇒Σ_(k=1) ^n (a_k ^2 +a_(k+3) ^2 −2a_k a_(k+3) )=0⇒Σ_(k=1) ^n (a_(k+3) −a_k )^2 ⇒∀k∈[0,n] a_(k+3) =a_k ⇒ 3∣n if periode 1 ⇒a_3 =a_2 =a_1 ⇒a_1 ^2 +1=a_1 since X^2 −X+1=0has non solution overR ⇒a_n can not constante for n=3 (−1,−1,2)=(a_1 ,a_2 ,a_3 ) worcks a_4 =−1=a_(3+1) =a_1 a_5 =−1.2+1=−1=a_(3+2) =a_2 the smalest periodicity is 3$
	$we can extend without$ $and U_{i + n} = a_{i}, \forall i \in [1, n]$ $U_{i} . U_{i + 1} + 1 = U_{i + 2}$ $\Rightarrow a_{n + 3} = a_{3}, if a_{1}, . . . . . . . a_{n}$ $the sequence U_{n} exist since is (a_{1,}, a_{2}, . . . . . a_{n}) is repeted infintly times$ $if U_{n} exist \Rightarrow a_{i} existe tack just n + 2 first therm$ $so we can extend$ $a_{i} without lost a_{i + 1} a_{i} + 1 = a_{i + 2}$ ${\begin{cases} a_{i + 2} a_{i + 1} + 1 = a_{i + 3} . . . . . . 1 \\ a_{i} a_{i + 1} + 1 = a_{i + 2} . . . 2 \end{cases}$ $1 * a_{i} - . . 2 * a_{i + 2} \Rightarrow a_{i} a_{i + 3} - a_{i + 2}^{2} = a_{i} - a_{i + 2}$ $\underset{i = 1}{\sum^{n}} (a_{i} a_{i + 3} - a_{i + 2}^{2}) = \underset{i = 1}{\sum^{n}} a_{i} - a_{i + 2} = a_{1} + a_{2} - a_{n + 2} - a_{n + 1} = 0$ $\Rightarrow \underset{i = 1}{\sum^{n}} (a_{i} a_{i + 3} - a_{i + 2}^{2}) = 0 . . . E$ $\underset{i = 1}{\sum^{n}} a_{i + 2} = \underset{k = 1}{\sum^{n}} a_{k}^{2}, since a_{n + 1} = a_{1} & a_{n + 2} = a_{2}$ $\Rightarrow E \Leftrightarrow \underset{k = 1}{\sum^{n}} (a_{k} a_{k + 3} - a_{k}^{2}) = 0$ $\underset{k = 1}{\sum^{n}} a_{k + 3}^{2} = \underset{k = 1}{\sum^{n}} a_{k}^{2} - a_{3}^{2} + a_{n + 3}^{2}, withe our expended series = \underset{k = 1}{\sum^{n}} a_{k}^{2}$ $\Rightarrow E \Leftrightarrow 2 \underset{k = 1}{\sum^{n}} a_{k} a_{k + 3} - 2 \underset{k = 1}{\sum^{n}} a_{k}^{2} = 0$ $\Rightarrow \underset{k = 1}{\sum^{n}} (a_{k}^{2} + a_{k + 3}^{2} - {2 a}_{k} a_{k + 3}) = 0 \Rightarrow \underset{k = 1}{\sum^{n}} {(a_{k + 3} - a_{k})}^{2}$ $\Rightarrow \forall k \in [0, n]$ $a_{k + 3} = a_{k} \Rightarrow 3 ∣ n$ $if periode 1$ $\Rightarrow a_{3} = a_{2} = a_{1} \Rightarrow a_{1}^{2} + 1 = a_{1} since X^{2} - X + 1 = 0 has non solution over R$ $\Rightarrow a_{n} can not constante$ $for n = 3$ $(- 1, - 1, 2) = (a_{1}, a_{2}, a_{3})$ $worcks$ $a_{4} = - 1 = a_{3 + 1} = a_{1}$ $a_{5} = - 1.2 + 1 = - 1 = a_{3 + 2} = a_{2}$ $the smalest periodicity is 3$
	Commented by chess1 last updated on 27/Nov/19

	$great$
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