<s>[INST] Why is every answer of $5^k - 2^k$ divisible by 3? We have the formula $$5^k - 2^k$$
I have noticed that every answer you get from this formula is divisible by 3. At least, I think so. Why is this? Does it have to do with $5-2=3$?
 [/INST] Let's use induction to prove that:
\begin{equation}
5^k - 2^k
\end{equation}
is divisible by 3
Let's check the proposition for k = 0 is divisible by 3
\begin{equation}
5^0 - 2^0 = 1 - 1 = 0
\end{equation}
which is divisible by 3
Let's assume that for k = n the proposition
\begin{equation}
5^n - 2^n
\end{equation}
is divisible by 3 and let's prove that
\begin{equation}
5^{(n+1)} - 2^{(n+1)}
\end{equation}
 is also true, so
\begin{equation}
    5^{(n+1)} - 2^{(n+1)} = 5{(5^n)} - 2{(2^n)}
\end{equation}
\begin{equation}
                      = (3 + 2)(5^n) - 2(2^n)
\end{equation}
\begin{equation}
                      = 3(5^n) + 2(5^n) - 2(2^n)
\end{equation}
\begin{equation}
                      = 3(5^n) + 2(5^n - 2^n)
\end{equation}
So in the above expression
\begin{equation}
    3(5^n)
\end{equation}
is divisible by 3
\begin{equation}
    2(5^n - 2^n)
\end{equation}
is divisible by 3 as per our assumption, so the finale expression
\begin{equation}
    3(5^n) + 2(5^n - 2^n)
\end{equation}
is also divisible by 3
So we can conclude that
\begin{equation}
    5^{(n+1)} - 2^{(n+1)}
\end{equation}
is divisible by 3
Meaning
\begin{equation}
    5^n - 2^n
\end{equation}
is divisible by 3 for
\begin{equation}
n \in \mathbb{N}
\end{equation}
</s>