show that 8^n −3^n is divisible by 5 for all natural number.


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Question Number 148732 by ethiork last updated on 30/Jul/21

$s h o w t h a t$ $8^{n} - 3^{n} i s d i v i s i b l e b y 5 f o r a l l n a t u r a l$ $n u m b e r .$
	Answered by Rasheed.Sindhi last updated on 30/Jul/21

	$5 ∣ (8^{n} - 3^{n})$ $⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣$ $∵ 8 \equiv 3 (m o d 5)$ $∴ 8^{n} \equiv 3^{n} (m o d 5)$
	Answered by mr W last updated on 30/Jul/21

	$8^{n} = {(5 + 3)}^{n} = 3^{n} + \underset{k = 1}{\sum^{n}} C_{k}^{n} 3^{n - k} 5^{k}$ $8^{n} - 3^{n} = \underset{k = 1}{\sum^{n}} C_{k}^{n} 3^{n - k} 5^{k} \equiv 0 m o d 5$
	Answered by physicstutes last updated on 31/Jul/21

	$let f (n) = 8^{n} - 3^{n}$ $f (1) = 8^{1} - 3^{1} = 5 = 5 (1) \Rightarrow true for f (1)$ $assume f (k) is true \Rightarrow 8^{k} - 3^{k} = 5 n, n \in N$ $now f (k + 1) = 8^{k + 1} - 3^{k + 1} = 8 (8^{k}) - 3 (3^{k})$ $f (k + 1) = 8 (5 n + 3^{k}) - 3 (3^{k})$ $= 5 (8 n) + 5 (3^{k})$ $= 5 (8 n + 3^{k}) since k \in N, 3^{k} \in N$ $and so (8 n + 3^{k}) \in N \Rightarrow f (k + 1) = 5 p$ $p \in N \Rightarrow f (k + 1) is true and so for all$ $natural numbers the expression f (n)$ $is divisible by 5 .$
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