Problem source

Let $n$ be a positive integer. 
\begin{questionparts}
\item Factorise $n^{5}-n^{3},$ and show that it is divisible by 24. 
\item Prove that $2^{2n}-1$ is divisible by 3. 
\item If $n-1$ is divisible by 3, show that $n^{3}-1$ is  divisible by 9. 
\end{questionparts}

Solution source

\begin{questionparts}
\item $n^5 -n^3 = n^3(n-1)(n+1)$. If $n$ is even then $8 \mid n^3$. if $n$ is odd then both of $n-1$ and $n+1$ are divisible by $2$ and one is divisible by $4$, so regardless $8$ divides our expression.

We can write $n = 3k, 3k+1, 3k+2$ and in all cases our expression is divisible by $3$. $n = 3k \Rightarrow 3 \mid n$, $n = 3k+1 \Rightarrow 3 \mid n-1$, $n = 3k+2 \Rightarrow 3 \mid n+1$.

Therefore $3$ and $8$ both divide our expression, and they are coprime so their product (24) divides our expression.

\item $2^{2n}-1 = (2^2-1) \cdot (1+2^2+\cdots + 2^{2n-2}) = 3 \cdot N$ therefore $3$ divides our number.

\item Suppose $n-1 = 3k$ then $n^3-1 = (3k+1)^3-1 = 27k^3 + 27k^2 + 9k$ which is clearly divisible by 9
\end{questionparts}