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Question Number 87878 by Cmr 237 last updated on 06/Apr/20
posons(1+23)n=an+bn3montrequepgcd(an;bn)=1
Answered by mind is power last updated on 07/Apr/20
an+1=an+6bnbn+1=bn+2an(an+1bn+1)=(1612)(anbn)M=(1612)Det(M−xI2)=0⇔(1−x)(2−x)−6=x2−3x−4=0⇒x∈{−1,4}{2x+6y=0x+3y=0⇒(x,y)=a(−3,1){−3x+6y=0x−2y=0⇔(x,y)=c(2,1)P=(−3211)D=(−1004)P−=1−5(1−2−1−3)Mn=−15(−3211).((−1)n004n)(1−2−1−3)Mn=−15(−3211).((−1)n−2(−1)n−4n−3.4n)=−15(3(−1)n+1−2.4n6(−1)n−6.4n(−1)n−4n2(−1)n+1−3.4n)(anbn)=Mn(a0b0)(1+23)0=1=a0+b03⇒a0=1,b0=0cause3∉Q(anbn)=−15(3(−1)n+1−2.4n(−1)n−4n)gcd(5an,5bn)∣(−3(−1)n−2.4n);(−1)n−4n)⇒gcd(5an,5bn)∣−5(−1)n⇔gcd(an,bn)∣(−1)n+1⇒gcd(an,bn)=1
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