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Question Number 160922 by KONE last updated on 09/Dec/21

monter que (3+(√5))^n +(3−(√5))^(n ) est divisible par 2^n beson d′aide svp

$${monter}\:{que}\:\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}\:} \\ $$$${est}\:{divisible}\:{par}\:\mathrm{2}^{{n}} \\ $$$${beson}\:{d}'{aide}\:{svp} \\ $$

Answered by mindispower last updated on 09/Dec/21

Starte withe U_(n+2) +aU_(n+1) +bU_n =0⇒U_n =c_1 r_1 ^n +c_2 r_2 ^2 r_1 ,r_2 roots of X^2 +aX+b=0 U_n =(3+(√5))^n +(3−(√5))^n ⇒ U_(n+2) =6U_(n+1) −4U_n U_0 =2,U_1 =6, lets prov that 2^n ∣U_(n, ) by induction n=0⇒2^0 =1∣2=U_0 ,n=1,2^1 =2∣6=U_1 ,suppose ∀n≥2 and for all m≤n 2^m ∣U_m , lets proof that 2^(n+1) ∣U_(n+1) U_(n+1) =6U_n −4U_(n−1) 2^n ∣U_n ,2^(n−1) ∣U_(n−1) U_n =q2^n ,U_(n−1) =w.2^(n−1) U_(n+1) =6.q.2^n −4.w.2^(n−1) =2^(n+1) (3q−w)⇒ 2^(n+1) ∣U_(n+1) ⇒∀n∈N,2^n ∣(3+(√5))^n +(3−(√5))^n

$${Starte}\:{withe} \\ $$$${U}_{{n}+\mathrm{2}} +{aU}_{{n}+\mathrm{1}} +{bU}_{{n}} =\mathrm{0}\Rightarrow{U}_{{n}} ={c}_{\mathrm{1}} {r}_{\mathrm{1}} ^{{n}} +{c}_{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} ,{r}_{\mathrm{2}} \:\:{roots}\:{of}\:{X}^{\mathrm{2}} +{aX}+{b}=\mathrm{0} \\ $$$${U}_{{n}} =\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}} \Rightarrow \\ $$$${U}_{{n}+\mathrm{2}} =\mathrm{6}{U}_{{n}+\mathrm{1}} −\mathrm{4}{U}_{{n}} \\ $$$${U}_{\mathrm{0}} =\mathrm{2},{U}_{\mathrm{1}} =\mathrm{6}, \\ $$$${lets}\:{prov}\:{that}\:\mathrm{2}^{{n}} \mid{U}_{{n},\:} {by}\:{induction} \\ $$$${n}=\mathrm{0}\Rightarrow\mathrm{2}^{\mathrm{0}} =\mathrm{1}\mid\mathrm{2}={U}_{\mathrm{0}} \:,{n}=\mathrm{1},\mathrm{2}^{\mathrm{1}} =\mathrm{2}\mid\mathrm{6}={U}_{\mathrm{1}} ,{suppose}\:\forall{n}\geqslant\mathrm{2}\:{and}\:{for}\:{all}\:{m}\leqslant{n} \\ $$$$\mathrm{2}^{{m}} \mid{U}_{{m}} , \\ $$$${lets}\:{proof}\:{that}\:\mathrm{2}^{{n}+\mathrm{1}} \mid{U}_{{n}+\mathrm{1}} \\ $$$${U}_{{n}+\mathrm{1}} =\mathrm{6}{U}_{{n}} −\mathrm{4}{U}_{{n}−\mathrm{1}} \\ $$$$\mathrm{2}^{{n}} \mid{U}_{{n}} ,\mathrm{2}^{{n}−\mathrm{1}} \mid{U}_{{n}−\mathrm{1}} \\ $$$${U}_{{n}} ={q}\mathrm{2}^{{n}} ,{U}_{{n}−\mathrm{1}} ={w}.\mathrm{2}^{{n}−\mathrm{1}} \\ $$$${U}_{{n}+\mathrm{1}} =\mathrm{6}.{q}.\mathrm{2}^{{n}} −\mathrm{4}.{w}.\mathrm{2}^{{n}−\mathrm{1}} =\mathrm{2}^{{n}+\mathrm{1}} \left(\mathrm{3}{q}−{w}\right)\Rightarrow \\ $$$$\mathrm{2}^{{n}+\mathrm{1}} \mid{U}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N},\mathrm{2}^{{n}} \mid\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}} \\ $$

Commented by KONE last updated on 14/Dec/21

merci merci bien que Dieu vlus benisse

$${merci}\:{merci}\:{bien}\:{que}\:{Dieu}\:{vlus}\:{benisse} \\ $$

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