Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $41^n - 14^n$ is a multiple of 27.

Let $P(n)$ be the statement: $41^n-14^n$ is a multiple of $27$.

Base case: For $n=1$,

a multiple of $27$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$, i.e. there is an integer $m$ with

We must show $41^{k+1}-14^{k+1}$ is a multiple of $27$.

Insert and subtract $41\cdot14^k$:

Substitute the induction hypothesis and simplify $41-14=27$: …

Divisibility induction using the "add and subtract $41\cdot14^k$" trick: this splits $41\cdot41^k-14\cdot14^k$ into $41(41^k-14^k)$ (uses the hypothesis) plus $14^ …

• Not spotting that $41-14=27$ is exactly the modulus needed, which is what makes the second piece $14^k(41-14)$ automatically a multiple of $27$. …