Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.

we have  $5^2 \equiv 1\ (mod\ 3) \Rightarrow 5^{2(3n+3)} \equiv 1\ (mod\ 3) \Rightarrow 5^{6n+6} \equiv 1\ (mod\ 3) $
and we have $2^2 \equiv 1\ (mod\ 3) \Rightarrow 2^{2(n+1)} \equiv 1\ (mod\ 3) \Rightarrow 2^{2n+2} \equiv 1\ (mod\ 3) $  Thus  $$5^{6n+6}-2^{2n+2} \equiv 0\ (mod\ 3)$$
