calculus - prove that for every natural n, $5^n - 2^n$, can be divided by 3

How to prove, using recursion, that for every natural n: $$5^n - 2^n$$

can be divided by 3.

Answer

1. setting $n=1$, $\implies 5^1-2^1=3$ is divisible by $3$

Thus, the number $5^n-2^n$ is divisible by $3$ for $n=1$

2. assume for $n=k$, the number $5^n-2^n$ is divisible by $3$ then $$\color{blue}{5^k-2^k=}\color{blue}{3m}$$
   
   where, $m$ is some integer




3. setting $n=k+1$, $$5^{k+1}-2^{k+1}=5\cdot 5^k-2\cdot 2^k$$
   $$=5\cdot 5^k-5\cdot 2^k+3\cdot 2^k$$
   $$=5(\color{blue}{5^k-2^k})+3\cdot 2^k$$
   $$=5(\color{blue}{3m})+3\cdot 2^k$$
   $$=3(5m+2^k)$$
   since, $(5m+2^k)$ is an integer hence, the above number $3(5m+2^k)$ is divisible by $3$

Hence, $5^n-2^n$ is divisible by $3$ for all integers $n\ge 1$