If \(\displaystyle{n}\in\mathbb{Z}\) is a positive ineger then \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}\) divisible by 17.

that is: \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}={x}^{{n}}{9}^{{n}}-{1}={18}^{{n}}-{1}\) is divisible by 17

For n = 1

18-1=17

17 is diviseble by 17

so, the statement is true for n=1

Let us assume that the given statement is true for n = k that is:

\(\displaystyle{k}\in{Z}\) is posisitve integer then \(\displaystyle{2}^{{k}}{3}^{{{2}{k}}}-{1}={18}^{{k}}-{1}\) is divisible by 17

Now, prove that the statement is true for n = k+1:

So, consider

\(\displaystyle{2}^{{{n}+{1}}}{3}^{{{2}{\left({n}+{1}\right)}}}-{1}={2}^{{n}}\cdot{2}\cdot{3}^{{{2}{n}}}-{1}\)

\(\displaystyle={18}\cdot{2}^{{n}}{3}^{{{2}{n}}}-{1}\)

\(\displaystyle={17}\cdot{2}^{{n}}{3}^{{{2}{n}}}+{2}^{{n}}{3}^{{{2}{n}}}-{1}\)

Now, since

\(\displaystyle{17}\cdot{2}^{{n}}{3}^{{{2}{n}}}\) is divisible by 17 and from assumption \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}\) is divisible by 17

hence, \(\displaystyle{2}^{{{n}+{1}}}{3}^{{{2}{\left({n}+{1}\right)}}}-{1}\) is diviseble by 17

Therefore, whenever n = k is true n = k+1 is true.

Therefore, it is true for all integers.