<s>[INST] Show that $9|2^{2n+1}+2^{4n+1}-4$ 
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
  1-Using induction
  2-Don't use induction  

I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
 [/INST] It's true for all non-negative integers $n$.
Without induction:
$9\mid 2^{2n+1}+2^{4n+1}-4$ is equivalent to $$9\mid 2\cdot 2^{2n+1}+\left(2^{2n+1}\right)^2-8$$
$$=\left(2^{2n+1}+1\right)^2-9$$
I.e. $9=3^2\mid \left(2^{2n+1}+1\right)^2$, i.e. $3\mid 2^{2n+1}+1$, which is true,
because $2^{2n+1}\equiv (-1)^{2n+1}\equiv -1\pmod{3}$, because $2n+1$ is odd for all $n\in\mathbb Z$, $n\ge 0$.

With induction:
Clearly $9\mid 2^{2\cdot 0+1}+2^{4\cdot 0+1}-4=0$.
If $9\mid 2^{2n+1}+2^{4n+1}-4$, then
$$9\mid 2^{2(n+1)+1}+2^{4(n+1)+1}-4$$
$$=\left(2^{2n+1}+2^{4n+1}-4\right)+3\left(2^{2n+1}+5\cdot 2^{4n+1}\right)$$
because $$2^{2n+1}+5\cdot 2^{4n+1}\pmod{3}$$
$$\equiv (-1)^{2n+1}-(-1)^{4n+1}\pmod{3}$$
$$\equiv -1-(-1)\equiv 0\pmod{3}$$
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