If n is a natural number, then $9^{2\mathrm{n}}-4^{2\mathrm{n}}$ is always divisible by

$9^{2\mathrm{n}}-4^{2\mathrm{n}}$ is of the form ${\mathrm{a}}^{2\mathrm{n}}-{\mathrm{b}}^{2\mathrm{n}}$which is divisible by both a - b and a + b .
So, $9^{2\mathrm{n}}-4^{2\mathrm{n}}$is divisible by both $9-4=5$ and $9+4=13$.