exercise Let a and b be natural integers such that 0<a<b. 1. Show that if a divides b, then for any naturel number n, n^a −1 divides n^b −1. 2. For any non−zero naturel number n, prove that the remainder of the euclidean division of n^b −1 by n^a −1 is n^r −1 where r is the remainder of the euclidean division of b by a. 3. For any non−zero naturel number n, show that gcd(n^b −1, n^a −1) = n^d −1 where d = gcd(b,c). by professor henderson^(−) .


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Question Number 144980 by henderson last updated on 01/Jul/21

$\underset{―}{exercise}$ $Let a and b be natural integers such that 0 < a < b .$ $1 . Show that if a divides b, then for any naturel$ $number n, n^{a} - 1 divides n^{b} - 1 .$ $2 . For any non - zero naturel number n, prove$ $that the remainder of the euclidean division of$ $n^{b} - 1 by n^{a} - 1 is n^{r} - 1 where r is the remainder$ $of the euclidean division of b by a .$ $3 . For any non - zero naturel number n, show$ $that g c d (n^{b} - 1, n^{a} - 1) = n^{d} - 1 where d = g c d (b, c) .$ $\overset{―}{b y p r o f e s s o r h e n d e r s o n} .$
	Answered by mindispower last updated on 01/Jul/21

	$a ∣ b \Rightarrow n^{a} - 1 ∣ n^{b} - 1$ $X^{c n} - 1 = (X^{c} - 1) (1 + X^{c} + X^{2 c} . . . . . + X^{. c (n - 1)})$ $a ∣ b \Rightarrow b = m a$ $n^{b} - 1 = n^{m a} - 1 = (n^{a} - 1) . (1 + n^{a} + . . . . . . . . + n^{a (m - 1)}) . u s i n g (1)$ $\Rightarrow n^{a} - 1 ∣ n^{b} - 1$ $b = m a + r$ $n^{b} - 1 = n^{m a + r} - 1 + n^{r} - n^{r}$ $n^{b} - 1 = n^{r} (n^{m a} - 1) + n^{r} - 1$ $n^{a - 1} ∣ n^{m a} - 1 . . . . . b y u s i n g a ∣ m a$ $\Rightarrow n^{b} - 1 \equiv (n^{r} - 1) m o d (n^{a} - 1)$ $s i n c e 0 ⩽ n^{r} - 1 < m i n (n^{a} - 1, n^{b} - 1)$ $n^{r} - 1 i s r e m a i n d e r o f d i v i s i o n$ $n^{b} - 1 b y n^{a} - 1$ $l e t d = g c d (a, b)$ $\Rightarrow n^{d} - 1 ∣ n^{a} - 1, n^{b} - 1 w i t h e 1$ $s u p o o s e m ∣ (n^{a} - 1, n^{b} - 1)$ $b > a$ $b = a m + r$ $\Rightarrow m ∣ n^{a} - 1, n^{r} - 1$ $b y r e c c u r a i o n o f e u c l i d$ $\Rightarrow m ∣ n^{d} - 1 \Rightarrow n^{d} - 1 > m$ $\Rightarrow \forall M \in N M ∣ (n^{a} - 1, n^{b} - 1) \Rightarrow M ∣ n^{d} - 1$ $n^{d} - 1 = g c d (n^{a} - 1, n^{b} - 1)$
	Commented bygreg_ed last updated on 01/Jul/21

	$sir, this line n^{a - 1} ∣ n^{m a} - 1 . . . . . b y u s i n g a ∣ m a {isn}^{'} t$ $n^{a} - 1 ∣ n^{m a} - 1 . . . . . b y u s i n g a ∣ m a ?$
	Commented bymindispower last updated on 03/Jul/21

	$y e s s i r$
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