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

$$\mathrm{Help}$$

Commented by York12 last updated on 08/Oct/24

Commented by Frix last updated on 08/Oct/24

$$\mathrm{Fibonacci}\mathrm{Sequences}:$$$$a_n=\frac{1}{2\sqrt{2}}({(3+2\sqrt{2})}^n-{(3-2\sqrt{2})}^n)$$$$a_1=0;a_2=2;a_n=-a_{n-2}+6a_{n-1}$$$$$$$$b_n=\frac{1}{2}({(3+2\sqrt{2})}^n+{(3-2\sqrt{2})}^n)$$$$b_1=1;b_2=3;b_n=-b_{n-2}+6b_{n-1}$$$$$$$$\Rightarrow \gcd =1$$

Commented by York12 last updated on 08/Oct/24

$$\mathrm{Thank}\mathrm{you}$$

Commented by York12 last updated on 08/Oct/24

$$\mathrm{TBH}\mathrm{sir}\mathrm{I}\mathrm{do}\mathrm{not}\mathrm{understand}\mathrm{why}{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{neccesarliy}\mathrm{means}\mathrm{that}\mathrm{the}gcd=1$$$$\mathrm{if}\mathrm{possible}\mathrm{can}\mathrm{you}\mathrm{elaborate}\mathrm{please}.$$

Commented by Frix last updated on 08/Oct/24

$$\mathrm{I}\mathrm{tried}\mathrm{the}\mathrm{following}:$$$$a_1=0;a_2=2$$$$b_1=1;b_2=1+2$$$$a_3=12;b_3=5+12$$$$a_4=70;b_4=29+70$$$$a_5=408;b_5=169+408$$$$...$$$$a_k;b_k=a_k+n$$$$\gcd (a_k;a_k+n)=q\Rightarrow q\mid a_k\wedge q\mid n$$$$n\mathrm{is}\mathrm{odd}\Rightarrow q\mathrm{is}\mathrm{odd}$$$$\mathrm{I}\mathrm{found}\mathrm{no}q\ne 1\mathrm{up}\mathrm{to}k=100$$

Commented by York12 last updated on 09/Oct/24

Commented by York12 last updated on 09/Oct/24

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