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certificate-x-1-x-2n-1-1-and-x-1-x-2n-1-1-All-established-2024-10-16-

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Question Number 212519 by MrGaster last updated on 16/Oct/24
certificate: (x−1)∣(x^(2n+1) −1)and(x+1)∣(x^(2n+1) +1) All established [2024.10.16]
certificate:(x−1)∣(x2n+1−1)and(x+1)∣(x2n+1+1)Allestablished[2024.10.16]
Answered by mathmax last updated on 19/Oct/24
x^(2n+1) −1 =(x−1)(1+x +x^2 +...x^(2n) ) so x−1 divise x^(2n+1) −1 montrons par recurrence que x^(2n+1) +1 est divisible par x+1 p_n (x)=x^(2n+1) +1 p_0 (x)=x+1 (vraie) supposons que x+1/p_n p_(n+1) (x)=x^(2n+3) +1 =x^(2n+1) x^2 +1 but p_n =q(x+1) =x^(2n+1) +1 ⇒x^(2n+1) =q(x+1)−1 ⇒ p_(n+1) =(q(x+1)−1)x^2 +1 =qx^2 (x+1)−(x^2 −1) =(x+1){qx^2 −(x−1)} donc p_(n+1) est divisible par x+1
x2n+1−1=(x−1)(1+x+x2+…x2n)sox−1divisex2n+1−1montronsparrecurrencequex2n+1+1estdivisibleparx+1pn(x)=x2n+1+1p0(x)=x+1(vraie)supposonsquex+1/pnpn+1(x)=x2n+3+1=x2n+1x2+1butpn=q(x+1)=x2n+1+1⇒x2n+1=q(x+1)−1⇒pn+1=(q(x+1)−1)x2+1=qx2(x+1)−(x2−1)=(x+1){qx2−(x−1)}doncpn+1estdivisibleparx+1

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