Question

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

Answer 1

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 \(\displaystyle{n}\in{N}\)