The n positive numbers x_1 ,x_2 ,...x_n where n≥3, satisfy x_1 =1+(1/x_2 ),x_2 =1+(1/x_3 ), ... , x_(n−1) =1+(1/x_n ) and x_n =1+(1/x_1 ). Show that x_1 =x_2 =x_3 =...=x


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Question Number 1777 by 112358 last updated on 22/Sep/15

$T h e n p o s i t i v e n u m b e r s x_{1}, x_{2}, . . . x_{n}$ $w h e r e n ⩾ 3, s a t i s f y$ $x_{1} = 1 + \frac{1}{x_{2}}, x_{2} = 1 + \frac{1}{x_{3}}, . . ., x_{n - 1} = 1 + \frac{1}{x_{n}}$ $a n d x_{n} = 1 + \frac{1}{x_{1}} .$ $S h o w t h a t x_{1} = x_{2} = x_{3} = . . . = x_{n} .$
Commented by Rasheed Soomro last updated on 26/Sep/15
$This problem can be translated in modulo form as follows: ′′ The n positive numbers x_0 , x_1 ,...,x_(n−1) where n≥3, satisfy x_(i (mod n)) =1+(1/x_((i +1) (mod n)) ) , i∈{0,1,2,...} Show that x_0 =x_1 =x_2 =...=x_(n−2) =x_(n−1) ′′ Let the above statment be denoted by P(n). Then by mathematical induction (1) P(n) should be proved for n=3. (2) P(k) is supposed to be true and P(k+1) should be proved true. (1) is easy but (2) is challenging.$
$T h i s p r o b l e m c a n b e t r a n s l a t e d i n m o d u l o f o r m$ $a s f o l l o w s :$ $^{″} T h e n p o s i t i v e n u m b e r s x_{0}, x_{1}, . . ., x_{n - 1}$ $w h e r e n ⩾ 3, s a t i s f y$ $x_{i (m o d n)} = 1 + \frac{1}{x_{(i + 1) (m o d n)}}, i \in {0, 1, 2, . . .}$ $S h o w t h a t$ $x_{0} = x_{1} = x_{2} = . . . = x_{n - 2} = x_{n - 1}^{″}$ $L e t t h e a b o v e s t a t m e n t b e d e n o t e d b y P (n) .$ $T h e n b y m a t h e m a t i c a l i n d u c t i o n$ $(1) P (n) s h o u l d b e p r o v e d f o r n = 3 .$ $(2) P (k) i s s u p p o s e d t o b e t r u e a n d P (k + 1) s h o u l d b e$ $p r o v e d t r u e .$ $(1) i s e a s y b u t (2) i s c h a l l e n g i n g .$
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