Question

Is $3^{2x+2}-8n-9$ is divisible by 8?

Answer 1

First, consider the value of $3^{(2n+2)}$-8n-9 for n = 1, it is $3^{(2+2)}$-8-9 = 81 - 17 = 64
Now assume $3^{(2n+2)}$-8n-9 is divisible by 8 for a value of n. With this assumption we test whether $3^{(2n+2)}$-8n-9 is divisible by 8 for n = n+1
$3^{\left(2\right(n+1)+2)}$-8(n+1)-9
$=3^{(2n+2+2)}$-8n- 8-9
$=3^{(2n+2)}$*9 - 8n - 9 - 8
$=3^{(2n+2)}$ - 8n - 9 + $3^{(2n+2)}$*8 - 8n - 8
$=3^{(2n+2)}$ - 8n - 9 + 8*($3^{(2n+2)}$ - n - 1)
This is clearly divisible by 8 as we have assumed $3^{(2n+2)}$ - 8n - 9 is divisible by 8 and 8*($3^{(2n+2)}$ - n - 1) has 8 as a factor.
This proves that $3^{(2n+2)}$-8n-9 is divisible by 8 for all natural values of n