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

$certificate :$ $(x - 1) ∣ (x^{2 n + 1} - 1) and (x + 1) ∣ (x^{2 n + 1} + 1)$ $All established$ $[2024.10.16]$
	Answered by mathmax last updated on 19/Oct/24
	$x^(2n+1) −1 =(x−1)(1+x +x^2 +...x^(2n) ) so x−1 divise x^(2n+1) −1 montrons par recurrence que x^(2n+1) +1 est divisible par x+1 p_n (x)=x^(2n+1) +1 p_0 (x)=x+1 (vraie) supposons que x+1/p_n p_(n+1) (x)=x^(2n+3) +1 =x^(2n+1) x^2 +1 but p_n =q(x+1) =x^(2n+1) +1 ⇒x^(2n+1) =q(x+1)−1 ⇒ p_(n+1) =(q(x+1)−1)x^2 +1 =qx^2 (x+1)−(x^2 −1) =(x+1){qx^2 −(x−1)} donc p_(n+1) est divisible par x+1$
	$x^{2 n + 1} - 1 = (x - 1) (1 + x + x^{2} + . . . x^{2 n}) s o$ $x - 1 d i v i s e x^{2 n + 1} - 1$ $m o n t r o n s p a r r e c u r r e n c e q u e$ $x^{2 n + 1} + 1 e s t d i v i s i b l e p a r x + 1$ $p_{n} (x) = x^{2 n + 1} + 1$ $p_{0} (x) = x + 1 (v r a i e) s u p p o s o n s q u e$ $x + 1 / p_{n}$ $p_{n + 1} (x) = x^{2 n + 3} + 1$ $= x^{2 n + 1} x^{2} + 1 b u t p_{n} = q (x + 1)$ $= x^{2 n + 1} + 1 \Rightarrow x^{2 n + 1} = q (x + 1) - 1 \Rightarrow$ $p_{n + 1} = (q (x + 1) - 1) x^{2} + 1$ $= {q x}^{2} (x + 1) - (x^{2} - 1)$ $= (x + 1) {{q x}^{2} - (x - 1)} d o n c p_{n + 1} e s t$ $d i v i s i b l e p a r x + 1$
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