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.

Well, $x-y$ divides $x^n-y^n$ so $41-14=27$ divides your expression. — lulu, Dec 31 '19 at 16:43
https://math.stackexchange.com/questions/188657/why-is-an-bn-divisible-by-a-b — lab bhattacharjee, Dec 31 '19 at 16:46

score 1 · Answer 1 · answered Dec 31 '19 at 16:44

1

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.

answered Dec 31 '19 at 16:44

Izaak van Dongen

3,791

score 0 · Accepted Answer · answered Dec 31 '19 at 17:10

0

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

answered Dec 31 '19 at 17:10

lhf

216,483

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

2 Answers2