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Question Number 211437 by necx122 last updated on 09/Sep/24

Commented by AlagaIbile last updated on 09/Sep/24

I posted the question on Cowbellpedia Whatshapp group two hours ago. How did it reach here?

$$\:\:{I}\:{posted}\:{the}\:{question}\:{on}\: \\ $$$$\:{Cowbellpedia}\:{Whatshapp}\:{group}\:\:{two} \\ $$$$\:{hours}\:{ago}.\:{How}\:{did}\:{it}\:{reach}\:{here}? \\ $$

Answered by A5T last updated on 09/Sep/24

x^2 =x+1⇒x^4 =x^2 +2x+1=(x+1)+2x+1=3x+2 ⇒x^8 =9x^2 +12x+4=9(x+1)+12x+4=21x+13 ⇒x^(16) =441(x^2 )+546x+169=987x+610 ⇒x^(20) =x^(16) ×x^4 =(987x+610)(3x+2) =2961x^2 +3804x+1220=6765x+4181=ax+b ⇒a+b=10946

$${x}^{\mathrm{2}} ={x}+\mathrm{1}\Rightarrow{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)+\mathrm{2}{x}+\mathrm{1}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{8}} =\mathrm{9}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{4}=\mathrm{9}\left({x}+\mathrm{1}\right)+\mathrm{12}{x}+\mathrm{4}=\mathrm{21}{x}+\mathrm{13} \\ $$$$\Rightarrow{x}^{\mathrm{16}} =\mathrm{441}\left({x}^{\mathrm{2}} \right)+\mathrm{546}{x}+\mathrm{169}=\mathrm{987}{x}+\mathrm{610} \\ $$$$\Rightarrow{x}^{\mathrm{20}} ={x}^{\mathrm{16}} ×{x}^{\mathrm{4}} =\left(\mathrm{987}{x}+\mathrm{610}\right)\left(\mathrm{3}{x}+\mathrm{2}\right) \\ $$$$=\mathrm{2961}{x}^{\mathrm{2}} +\mathrm{3804}{x}+\mathrm{1220}=\mathrm{6765}{x}+\mathrm{4181}={ax}+{b} \\ $$$$\Rightarrow{a}+{b}=\mathrm{10946} \\ $$

Answered by Rasheed.Sindhi last updated on 09/Sep/24

x^2 =x+1⇒x^4 =x^2 +2x+1 =(x+1)+2x+1=3x+2 ⇒x^5 =3x^2 +2x=3(x+1)+2x=5x+3 x^(20) =(x^5 )^4 ={(5x+3)^2 }^2 =(25x^2 +30x+9)^2 ={25(x+1)+30x+9}^2 =(55x+34)^2 =3025x^2 +3740x+1156 =3025(x+1)+3740x+1156 =6765x+4181 a=6765 ,b=4181 a+b=10946

$${x}^{\mathrm{2}} ={x}+\mathrm{1}\Rightarrow{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$$\:=\left({x}+\mathrm{1}\right)+\mathrm{2}{x}+\mathrm{1}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{5}} =\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{3}\left({x}+\mathrm{1}\right)+\mathrm{2}{x}=\mathrm{5}{x}+\mathrm{3} \\ $$$$\: \\ $$$${x}^{\mathrm{20}} =\left({x}^{\mathrm{5}} \right)^{\mathrm{4}} =\left\{\left(\mathrm{5}{x}+\mathrm{3}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$=\left(\mathrm{25}{x}^{\mathrm{2}} +\mathrm{30}{x}+\mathrm{9}\right)^{\mathrm{2}} \\ $$$$=\left\{\mathrm{25}\left({x}+\mathrm{1}\right)+\mathrm{30}{x}+\mathrm{9}\right\}^{\mathrm{2}} \\ $$$$=\left(\mathrm{55}{x}+\mathrm{34}\right)^{\mathrm{2}} \\ $$$$=\mathrm{3025}{x}^{\mathrm{2}} +\mathrm{3740}{x}+\mathrm{1156} \\ $$$$=\mathrm{3025}\left({x}+\mathrm{1}\right)+\mathrm{3740}{x}+\mathrm{1156} \\ $$$$=\mathrm{6765}{x}+\mathrm{4181} \\ $$$${a}=\mathrm{6765}\:,{b}=\mathrm{4181} \\ $$$${a}+{b}=\mathrm{10946} \\ $$

Commented by necx122 last updated on 09/Sep/24

Thid id great

$${Thid}\:{id}\:{great} \\ $$

Answered by A5T last updated on 09/Sep/24

Observe that x^2 −x−1=0 is the characteristic equation of f_n =f_(n−1) +f_(n−2) ⇒x^n =f_n x+f_(n−1) ⇒x^(20) =f_(20) x+f_(19) ⇒a+b=f_(20) +f_(19) =f_(21) =10946 f_0 =0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610, 987,1597,2584,4181,6765,f_(21) =10946

$${Observe}\:{that}\:{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0}\:{is}\:{the}\:{characteristic} \\ $$$${equation}\:{of}\:{f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} \\ $$$$\Rightarrow{x}^{{n}} ={f}_{{n}} {x}+{f}_{{n}−\mathrm{1}} \Rightarrow{x}^{\mathrm{20}} ={f}_{\mathrm{20}} {x}+{f}_{\mathrm{19}} \\ $$$$\Rightarrow{a}+{b}={f}_{\mathrm{20}} +{f}_{\mathrm{19}} ={f}_{\mathrm{21}} =\mathrm{10946} \\ $$$${f}_{\mathrm{0}} =\mathrm{0},\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8},\mathrm{13},\mathrm{21},\mathrm{34},\mathrm{55},\mathrm{89},\mathrm{144},\mathrm{233},\mathrm{377},\mathrm{610}, \\ $$$$\mathrm{987},\mathrm{1597},\mathrm{2584},\mathrm{4181},\mathrm{6765},{f}_{\mathrm{21}} =\mathrm{10946} \\ $$

Commented by necx122 last updated on 09/Sep/24

hmmm.. never knew this. Cool stuff.

$${hmmm}..\:{never}\:{knew}\:{this}.\:{Cool}\:{stuff}. \\ $$

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