Prove that \(A_n=3^{\,n+3}-4^{\,4n+2}\) is divisible by \(11\) for every positive integer \(n\).

We will prove by Mathematical Induction.

For \(n=0\), \(A_0=3^3-4^2=27-16=11\), which is divisible by \(11\).

Assume that \(A_k\) is divisible by \(11\), i.e., \(11\mid A_k\).

We shall prove that \(A_{k+1}\) is also divisible by \(11\).

\[ \begin{aligned} A_{k+1} &=3^{k+4}-4^{4k+6}\\ &=3\cdot3^{k+3}-4^4\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-256\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-3\cdot4^{4k+2}-253\cdot4^{4k+2}\\ &=3\left(3^{k+3}-4^{4k+2}\right)-253\cdot4^{4k+2}\\ &=3A_k-253\cdot4^{4k+2}. \end{aligned} \]

Since \(11\mid A_k\) by the induction hypothesis and \(253=11\times23\), both terms on the right-hand side are divisible by \(11\). Hence \(11\mid A_{k+1}\).

Therefore, by the Principle of Mathematical Induction, \(11\mid\left(3^{n+3}-4^{4n+2}\right)\) for every nonnegative integer \(n\).

Solution by: Thin Sokkean