Mathematical Induction divisibility $8\mid 3^{2n}-1$

elementary-number-theory divisibility discrete-mathematics induction

Solution 1:

Hint: Note that: $$3^{2(n+1)} -1=9\cdot3^{2n}-1=(3^{2n}-1)+8\cdot3^{2n}$$

Solution 2:

If you don't mind I'm gonna do some magic $$3^{2n}=(3^2)^n=(8+1)^n=\binom{n}{o}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8^{1}+\binom{n}{n}8^0$$

$$3^{2n}=\binom{n}{0}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8^{1}+1$$

$$3^{2n}-1=\binom{n}{0}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8=8(k)$$

$$\implies8|(3^{2n} - 1)$$

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