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Divisibility by 133

Problem

Show by mathematical induction that for all \(n \in \mathbb{N}\) there exists an \(m \in \mathbb{N}\) such that the following equations holds.

\[ 11^{n+1} + 12^{2n-1} = 133 \cdot m \]

Proof

Initial Case: \(n=1\)

\[ 11^{n+1} + 12^{2n-1} = 11^2 + 12 = 133 \quad \implies \quad m = 1 \]

Induction Step: \(n\implies n+1\)

Let \(n\in\mathbb{N}\). According to the induction hypothesis, there exists an \(m\in\mathbb{N}\) such that the following holds.

\[ 11^{n+1} + 12^{2n-1} = 133 \cdot m \]

We will now show that based on this hypothesis there exists another \(m' \in \mathbb{N}\) such that the same equation holds for \(n+1\).

\[ 11^{(n+1)+1} + 12^{2(n+1)-1} = 11 \cdot 11^{n+1} + 12^2 \cdot 12^{2n-1} \]

First, we factor out the number \(11\). For the right term, we are left with a remaining part.

\[ = 11 \cdot \left( 11^{n+1} + 12^{2n-1} \right) + \left( 12^2 - 11 \right) \cdot 12^{2n-1} \]
\[ = 11 \cdot \left( 11^{n+1} + 12^{2n-1} \right) + 133 \cdot 12^{2n-1} \]

Using the induction hypothesis, we know that the first term is divisible by \(133\). After factoring out \(133\), it can then be seen that for \(n+1\) the given number is divisible by \(133\).

\[ = 11 \cdot 133 \cdot m + 133 \cdot 12^{2n-1} \]
\[ = 133 \cdot \left( 11 \cdot m + 12^{2n-1} \right) \quad\implies\quad m' = 11 \cdot m + 12^{2n-1} \]

By mathematical induction, this proves the lemma.

Last update: August 10, 2020