For every positive integer n, prove that 7^{n | vedclass

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(N/A) We define the statement $P(n): 7^{n} - 3^{n}$ is divisible by $4.$Step $1$: Check for $n = 1.$$P(1): 7^{1} - 3^{1} = 4,$ which is divisible by $

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(N/A) We define the statement $P(n): 7^{n} - 3^{n}$ is divisible by $4.$
Step $1$: Check for $n = 1.$
$P(1): 7^{1} - 3^{1} = 4,$ which is divisible by $4.$ Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k.$
$P(k): 7^{k} - 3^{k} = 4d$ for some integer $d \in \mathbb{N}.$
Step $3$: Prove $P(k+1)$ is true.
$7^{k+1} - 3^{k+1} = 7 \cdot 7^{k} - 3 \cdot 3^{k}$
$= 7 \cdot 7^{k} - 7 \cdot 3^{k} + 7 \cdot 3^{k} - 3 \cdot 3^{k}$
$= 7(7^{k} - 3^{k}) + (7 - 3) \cdot 3^{k}$
$= 7(4d) + 4 \cdot 3^{k}$
$= 4(7d + 3^{k})$
Since $4(7d + 3^{k})$ is a multiple of $4,$ $P(k+1)$ is true.
By the principle of mathematical induction,$7^{n} - 3^{n}$ is divisible by $4$ for every positive integer $n.$

For every positive integer $n,$ prove that $7^{n}-3^{n}$ is divisible by $4.$

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