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Question Number 212242 by York12 last updated on 08/Oct/24

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Commented by York12 last updated on 08/Oct/24

Commented by Frix last updated on 08/Oct/24

Fibonacci Sequences :

a_{n} = \frac{1}{2 \sqrt{2}} ({(3 + 2 \sqrt{2})}^{n} - {(3 - 2 \sqrt{2})}^{n})

a_{1} = 0; a_{2} = 2; a_{n} = - a_{n - 2} + 6 a_{n - 1}

b_{n} = \frac{1}{2} ({(3 + 2 \sqrt{2})}^{n} + {(3 - 2 \sqrt{2})}^{n})

b_{1} = 1; b_{2} = 3; b_{n} = - b_{n - 2} + 6 b_{n - 1}

\Rightarrow \gcd = 1

Commented by York12 last updated on 08/Oct/24

Thank you

Commented by York12 last updated on 08/Oct/24

TBH sir I do not understand why {it}^{'} s

neccesarliy means that the g c d = 1

if possible can you elaborate please .

Commented by Frix last updated on 08/Oct/24

I tried the following :

a_{1} = 0; a_{2} = 2

b_{1} = 1; b_{2} = 1 + 2

a_{3} = 12; b_{3} = 5 + 12

a_{4} = 70; b_{4} = 29 + 70

a_{5} = 408; b_{5} = 169 + 408

\dots

a_{k}; b_{k} = a_{k} + n

\gcd (a_{k}; a_{k} + n) = q \Rightarrow q ∣ a_{k} \land q ∣ n

n is odd \Rightarrow q is odd

I found no q \neq 1 up to k = 100

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