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 $3\mid(5^{6n+1}-2^{2n+1})$.

\begin{align}
5^{6n+6}-2^{2n+2} &\equiv 2^{6n+6}-2^{2n+2} \pmod{3} \\
&\equiv  2^{2n+2}(2^{(2n+2)2}-1) \pmod{3}\\
&\equiv2^{2n+2}(1-1) \pmod{3}\\
&\equiv 0 \pmod{3}
\end{align}
Notice that I used  $\gcd(3, 2^{2n+2})=1$ and Fermat's theorem to conclude that $2^{(2n+2)2}\equiv1 \pmod{3}$.
