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

Oct 19, 2016

See below.

#### Explanation:

For $n = 1$ we have $f \left(1\right) = 8$ which is divisible by $8$
Now, supposing that $f \left(n\right)$ is divisible by $8$ then

$f \left(n\right) = {2}^{2} + {6}^{2} = 8 \cdot k$

The last step is verify if $f \left(n + 1\right)$ is divisible by $8$.

$f \left(n + 1\right) = {2}^{n + 1} + {6}^{n + 1} = 2 \cdot {2}^{n} + 6 \cdot {6}^{n} = 6 \cdot {2}^{n} + 6 \cdot {6}^{n} - 4 \cdot {2}^{n} = 6 \cdot 8 \cdot k - {2}^{2} \cdot {2}^{n}$

but if $n > 1$ we have

$f \left(n + 1\right) = 6 \cdot 8 \cdot k - {2}^{2} \cdot {2}^{n}$ is divisible by $8$

so inductively the assertion is true.