202210201121 Solution to 1977-CE-AMATH-I-XX

Prove, by mathematical induction, that $23^n-1$ is divisible by $11$ .

Roughwork.

Let proposition $P(n)$ be that

$P(n):\enspace 11|23^n-1$ .

First, note

$\begin{aligned} & \quad\enspace P(1):\enspace 11|23^{1}-1 \\ & \Leftrightarrow P(1):\enspace 11|23-1 \\ & \Leftrightarrow P(1):\enspace 11|22 \\ & \Leftrightarrow P(1)\textrm{ is true} \\ \end{aligned}$

that $P(n)$ is true for $n=1$ .

Next, assumed be to $P(n)$ true.

$\begin{aligned} &\quad\enspace P(n+1):\enspace 11|23^{(n+1)}-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^{n}\cdot 23-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^n-1+23^n\cdot 22\\ & \dots\textrm{ by assumption }11|23^n-1\textrm{ and for }11|22 \\ & \Leftrightarrow P(n+1)\textrm{ is true} \\ \end{aligned}$

Thus, $P(n)$ is true implies $P(n+1)$ true.

From $P(1)\textrm{ true}\,\vee\, P(n)\Longrightarrow P(n+1)\textrm{ true}$ follows $P(n)$ the proposition is true for all positive integers $n\geqslant 1$ . By the principle of mathematical induction proven is that $11$ divides $23^n-1$ .

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