Proof of the following statement : $41^a - 14^a$ is a multiple of $27$ for $a \in \Bbb N$

Question

Can anyone give me a simple proof of this statement . I know how to prove it using principle of mathematical induction so please prove it in some other way.

Answer 1

By the binomial theorem, $41^a = (27+14)^a = 27m + 14^a$, for some $m \in \mathbb N$.

Answer 2

Reducing modulo $27$, \begin{align*} 41^a - 14^a &\equiv 14^a - 14^a \\ &\equiv 0 \pmod{27} \end{align*} (since $41 = 27 + 14$) so $27$ divides your number.