<s>[INST] 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 $3\mid(5^{6n+1}-2^{2n+1})$.
 [/INST] By binomial theorem$$5^{6n+6}-2^{2n+2}$$
$$={(3+2)}^{6n+6}-2^{2n+2}$$
$$=3k+2^{6n+6}-2^{2n+2}$$
$$=3k+2^{2n+2}({(3-1)}^{4n+4}-1)=3k+2^{2n+2}(3m)$$
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