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Question Number 87878 by Cmr 237 last updated on 06/Apr/20

posons   (1+2(√3))^n =a_n +b_n (√3)  montre que pgcd(a_n ;b_n )=1

posons(1+23)n=an+bn3montrequepgcd(an;bn)=1

Answered by mind is power last updated on 07/Apr/20

a_(n+1) =a_n +6b_n   b_(n+1) =b_n +2a_n    ((a_(n+1) ),(b_(n+1) ) )= (((1     6)),((1     2)) ) ((a_n ),(b_n ) )  M= (((1      6)),((1      2)) )  Det(M−xI_2 )=0⇔(1−x)(2−x)−6  =x^2 −3x−4=0  ⇒x∈{−1,4}   { ((2x+6y=0)),((x+3y=0)) :}⇒(x,y)=a(−3,1)   { ((−3x+6y=0)),((x−2y=0)) :}⇔(x,y)=c(2,1)  P= (((−3      2)),((   1       1 )) )   D= (((−1       0)),((   0         4)) )  P^− =(1/(−5)) (((  1          −2)),((−1      −3)) )   M^n =−(1/5) (((−3      2)),((1          1)) ). ((((−1)^n       0)),((   0             4^n )) ) (((1       −2)),((−1      −3)) )  M^n =−(1/5) (((−3     2)),((1         1)) ). ((((−1)^n       −2(−1)^n )),((−4^n            −3.4^n )) )  =((−1)/5) (((3(−1)^(n+1) −2.4^n       6(−1)^n −6.4^n )),(((−1)^n −4^n                 2(−1)^(n+1) −3.4^n )) )   ((a_n ),(b_n ) )=M^n  ((a_0 ),(b_0 ) )  (1+2(√3))^0 =1=a_0 +b_0 (√3)⇒a_0 =1,b_0 =0  cause(√3)∉Q   ((a_n ),(b_n ) )=−(1/5) (((3(−1)^(n+1) −2.4^n )),(((−1)^n −4^n )) )  gcd(5a_n ,5b_n )∣(−3(−1)^n −2.4^n );(−1)^n −4^n )  ⇒gcd(5a_n ,5b_n )∣−5(−1)^n ⇔gcd(a_n ,b_n )∣(−1)^(n+1)   ⇒gcd(a_n ,b_n )=1

an+1=an+6bnbn+1=bn+2an(an+1bn+1)=(1612)(anbn)M=(1612)Det(MxI2)=0(1x)(2x)6=x23x4=0x{1,4}{2x+6y=0x+3y=0(x,y)=a(3,1){3x+6y=0x2y=0(x,y)=c(2,1)P=(3211)D=(1004)P=15(1213)Mn=15(3211).((1)n004n)(1213)Mn=15(3211).((1)n2(1)n4n3.4n)=15(3(1)n+12.4n6(1)n6.4n(1)n4n2(1)n+13.4n)(anbn)=Mn(a0b0)(1+23)0=1=a0+b03a0=1,b0=0cause3Q(anbn)=15(3(1)n+12.4n(1)n4n)gcd(5an,5bn)(3(1)n2.4n);(1)n4n)gcd(5an,5bn)5(1)ngcd(an,bn)(1)n+1gcd(an,bn)=1

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