Is $7^{6n}-6^{6n}$ always divisible by $13$,$127$ and $559$, for any natural number $n$?
$7^{6n}-6^{6n}={(7^{3n})}^{2}-{(6^{3n})}^{2}$
${(7^{3n})}^{2}-{(6^{3n})}^{2}=(7^{3n}+6^{3n})(7^{3n}-6^{3n})$
$(7^{3n}+6^{3n})(7^{3n}-6^{3n})=(7^n+6^n)(7^{2n}-7^n\times 6^n+6^{2n})(7^n-6^n)(7^{2n}+7^n\times 6^n+6^{2n})$
when, $n=1$
$7^{6}-6^{6}=(7+6)(7^2-7\times 6+6^2)(7-6)(7^2+7\times 6+6^2)=(13)(43)(127)$