<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] $5^{6n}-2^{2n}=125^{2n}-2^{2n}$
$=(125-2)\left(125^{2n-1}+125^{2n-2}2+125^{2n-3}2^2+\cdots+125^22^{2n-3}+125\cdot2^{2n-2}+2^{2n-1}\right)$
is divisible by $3$, because $125-2$ is.
Now substitute $m=n+1$ to show that $5^{6m+6}-2^{2m+2}$ is divisible by $3$.
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