Modular Number Theory

Let the equation be true for n equal to some non-negative integer a. Note that it is true for n=0.

$3^{2a+2}-8a-9\equiv 0 \pmod{64}$

multiply through by 9

$3^{2a+4}-72a-81\equiv 0 \pmod{64}$

$3^{2a+4}-8a-8-9\equiv 64a+64 \pmod{64}$ (multiples of 64 are of course 0 (mod 64))

$3^{2a+4}-8a-8-9\equiv 0 \pmod{64}$

$3^{2(a+1)+2}-8(a+1)-9\equiv 0 \pmod{64}$

Thus if the equation is true for any n=a it is also true for n=a+1. It is true for n=0. Proof by induction.

Similar solution with @David Vreken 's

$\begin{aligned} N & = 3^{2n+2} - 8n - 9 \\ & = 9^{n+1} - 8n - 8 - 1 \\ & = {\color{#3D99F6}(8+1)^{n+1}} - 8(n+1) - 1 \\ & = {\color{#3D99F6}\sum_{k=0}^{n+1}\binom {n+1}k 8^k} - 8(n+1) - 1 \\ & = {\color{#3D99F6}\sum_{\color{#D61F06}k=2}^{n+1}\binom {n+1}k 8^k + 8(n+1)+1} - 8(n+1) - 1 \\ & = \sum_{\color{#D61F06}k=2}^{n+1}\binom {n+1}k 8^k \\ & = 64\sum_{\color{#3D99F6}k=0}^{n+1}\binom {n+1}{\color{#3D99F6}k+2} 8^k \\ & \equiv \boxed{\text{0 (mod 64)}} \end{aligned}$

Yes , it is always true.

$3^{2n + 2} - 8n - 9$

$= 3^{2(n + 1)} - 8n - 9$

$= 9^{n + 1} - 8n - 9$

$= (8 + 1)^{n + 1} - 8n - 8 - 1$

$= (8 + 1)^{n + 1} - 8(n + 1) - 1$

$= ({{n + 1} \choose {n + 1}}8^{n + 1}1^{0} + {{n + 1} \choose {n}}8^{n}1^{1} + {{n + 1} \choose {n - 1}}8^{n - 1}1^{2} + ... + {{n + 1} \choose {2}}8^{2}1^{n - 1} + {{n + 1} \choose {1}}8^{1}1^{n} + {{n + 1} \choose {0}}8^{0}1^{n + 1}) - 8(n + 1) - 1$ (by the binomial expansion)

$= (k8^2 + (n + 1)8 + 1) - 8(n + 1) - 1$ (for some integer $k$ )

$= k8^2$

$= 64k$

Since $64k$ is always divisible by $64$ , $3^{2n + 2} - 8n - 9 \equiv 0 \pmod{64}$ is always true .