<s>[INST] Showing that $2^6$ divides $3^{2264}-3^{104}$ 
Show that $3^{2264}-3^{104}$ is divisible by $2^6$. 

My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$. 
I have no idea how to show divisibility by $2^6$.
 [/INST] Another approach:
$$\text{Euler's formula:  }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$\implies a^{32}\equiv 1 \pmod{2^6}$
We want $3^{2264}-3^{104}\equiv 0 \pmod{2^6}$
$$\phi(2^6)=2^5=32$$
$$ 3^{2264}-3^{104}=3^{2240}3^{24}-3^{96}3^{8}=(3^{70})^{32}3^{24}-(3^3)^{32}3^8\equiv 3^{24}-3^8\equiv 17^6-17^2\\
\equiv 33^3-33\equiv \color{green}{0 \pmod{2^6}}$$
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