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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]
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\left({x}−\mathrm{1}\right)\mid\left({x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}\right)\mathrm{and}\left({x}+\mathrm{1}\right)\mid\left({x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\right) \\ $$$$\mathrm{All}\:\mathrm{established} \\ $$$$\left[\mathrm{2024}.\mathrm{10}.\mathrm{16}\right] \\ $$
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
$${x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left(\mathrm{1}+{x}\:+{x}^{\mathrm{2}} +…{x}^{\mathrm{2}{n}} \right)\:{so} \\ $$$${x}−\mathrm{1}\:{divise}\:{x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1} \\ $$$${montrons}\:{par}\:{recurrence}\:{que} \\ $$$${x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\:{est}\:{divisible}\:{par}\:{x}+\mathrm{1} \\ $$$${p}_{{n}} \left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1} \\ $$$${p}_{\mathrm{0}} \left({x}\right)={x}+\mathrm{1}\:\:\:\left({vraie}\right)\:{supposons}\:{que}\: \\ $$$${x}+\mathrm{1}/{p}_{{n}} \\ $$$${p}_{{n}+\mathrm{1}} \left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{3}} +\mathrm{1} \\ $$$$={x}^{\mathrm{2}{n}+\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{1}\:{but}\:{p}_{{n}} ={q}\left({x}+\mathrm{1}\right) \\ $$$$={x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}{n}+\mathrm{1}} ={q}\left({x}+\mathrm{1}\right)−\mathrm{1}\:\Rightarrow \\ $$$${p}_{{n}+\mathrm{1}} =\left({q}\left({x}+\mathrm{1}\right)−\mathrm{1}\right){x}^{\mathrm{2}} +\mathrm{1} \\ $$$$={qx}^{\mathrm{2}} \left({x}+\mathrm{1}\right)−\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$=\left({x}+\mathrm{1}\right)\left\{{qx}^{\mathrm{2}} −\left({x}−\mathrm{1}\right)\right\}\:{donc}\:{p}_{{n}+\mathrm{1}} {est} \\ $$$${divisible}\:{par}\:{x}+\mathrm{1} \\ $$

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