Question

Demonstrate that `2^n+6^n` is divisible by `8` for `n=1,2,3,cdots` ?

Answer 1

See below.

Explanation:

For `n=1` we have `f(1)=8` which is divisible by `8`
Now, supposing that `f(n)` is divisible by `8` then

`f(n) = 2^2+6^2=8 cdot k`

The last step is verify if `f(n+1)` is divisible by `8`.

`f(n+1) = 2^(n+1)+6^(n+1) = 2 cdot 2^n+6 cdot 6^n = 6 cdot 2^n + 6 cdot 6^n - 4 cdot 2^n = 6 cdot 8 cdot k-2^2cdot 2^n`

but if `n > 1` we have

`f(n+1)= 6 cdot 8 cdot k-2^2cdot 2^n` is divisible by `8`

so inductively the assertion is true.