Question

If $n$ is a positive integer$,$ prove that $3^{3n}-26n-1$ is divisible by $676.$

Answer 1

Factors of $676$ are $2\times 2\times 13\times 13$

For $n=1$ we have $3^3-26-1=27-26-1=0$ is divisible by $676$

For $n=2$ we have $3^6-26\times 2-1=729-52-1=676$ is divisible by $676$

For $n=n$ we have $3^{3n}-26n-1=729-52-1=676$ is divisible by $676$

For $n=n+1$ we have $3^{3(n+1)}-26(n+1)-1$

$=27.3^{3n}-26n-26-1$ ......$\left(1\right)$

Let $3^{3n}-26n-1=676k$ where $k$ is a constant.

$\Rightarrow 3^{3n}=676k+1+26n$

Substituting in $\left(1\right)$ we get

$27.3^{3n}-26n-26-1$

$\Rightarrow 27.(676k+1+26n)-26n-27$

$\Rightarrow 676\left(27k\right)+n(27\times 26-26)$

$\Rightarrow 676\left(27k\right)+676n$

$\Rightarrow 676(27k+n)$ is divisible by $676$

Hence proved.