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

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

2021-03-02
Claim: $$\displaystyle{3}^{{{2}{n}}}-{1}$$ is divisible by 4 for $$\displaystyle{n}\in{N}$$.
We will prove this by induction.
Base step: n = 1
$$\displaystyle\therefore{3}^{{2}}-{1}={8}$$, which is divisible by 4
$$\displaystyle\therefore$$the result is true fir n =1.
Inductive step:
Assure that result is true for n = k
i.e. $$\displaystyle\frac{{4}}{{3}^{{{2}{k}}}}-{1}$$...(1)
Now, we prove that: result is true form = k+1
i.e. $$\displaystyle\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}$$
Consider $$\displaystyle{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{3}^{{{2}{k}}}\cdot{9}{\left(\text{mod}{4}\right)}$$
$$\displaystyle\equiv{1}\cdot{9}{\left(\text{mod}{4}\right)}{\left[{\mathfrak{{o}}}{m}{\left({i}\right)}\right]}$$
$$\displaystyle\therefore{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{1}{\left(\text{mod}{4}\right)}$$
$$\displaystyle\therefore\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}$$
$$\displaystyle\therefore$$The result is true for n = k+1.
$$\displaystyle\therefore$$By induction,
$$\displaystyle{3}^{{{2}{\left({n}+{1}\right)}}}-{1}$$ is divisibly by 4 for $$\displaystyle{n}\in{N}$$

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