Question

Prove that the $8^{th}$ power of any number is of the form $17n$ or $17n\pm 1$.

Answer 1

If $N$ is prime to $17$ then

by using Femats little theorem if $p$ is a prime number and $N$ is prime to $p$ then $N^{p-1}-1$ is a multiple of $p$

Put $p=17$

$\Rightarrow N^{17-1}-1=17n\Rightarrow N^{16}-1=17n{\Rightarrow {\left(N^8\right)}^2-1}^2=17n\Rightarrow (N^8+1)(N^8-1)=17n\Rightarrow N^8=17n\pm 1$

If $N$ is not prime to $17$ then obviously

$N^8=17n$

Hence proved.