Answer
See below.
Work Step by Step
Proofs using mathematical induction consist of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number. Then, we prove that the statement also holds for $n + 1$.
Hence, here we have:
1) For $n=1: 8^1-3^1=5$ is divisible by $5$.
2) Assume for $n=k: 8^k-3^k$ is divisible by $5$. Then for $n=k+1$:
$8^{k+1}-3^{k+1}=8^k\cdot8-3^k\cdot3=8(8^k-3^k)+5\cdot 3^k$
We know from the inductive hypothesis above that $8^k-3^k$ is divisible by $5$. Multiplying this value by $8$ will not alter its divisibility by $5$. Similarly, adding $5*3^k$ will not alter its divisibility either because we are adding an integer number of $5$'s. Thus, the result is divisible by $5$ and we have proved what we wanted to.
