2021-03-01

Using the Mathematical Induction to prove that: ${3}^{2n}-1$ is divisible by 4, whenever n is a positive integer.

Jozlyn

Claim: ${3}^{2n}-1$ is divisible by 4 for $n\in N$.
We will prove this by induction.
Base step: $n=1$
$\therefore {3}^{2}-1=8$, which is divisible by 4
$\therefore$the result is true fir $n=1.$
Inductive step:
Assure that result is true for $n=k$
i.e. $\frac{4}{{3}^{2k}}-1$...(1)
Now, we prove that: result is true form $=k+1$
i.e. $\frac{4}{{3}^{2\left(k+1\right)}}-1$
Consider ${3}^{2\left(k+1\right)}\equiv {3}^{2k}\cdot 9\left(\text{mod}4\right)$
$\equiv 1\cdot 9\left(\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}4\right)\left[\mathfrak{o}m\left(i\right)\right]$
$\therefore {3}^{2\left(k+1\right)}\equiv 1\left(\text{mod}4\right)$
$\therefore \frac{4}{{3}^{2\left(k+1\right)}}-1$
$\therefore$The result is true for $n=k+1$.
$\therefore$By induction,
${3}^{2\left(n+1\right)}-1$ is divisibly by 4 for $n\in N$

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