Question

Prove that ${11}^{n+2}+{12}^{2n+1}$ is divisible by $133$ for any non-negative integral $n$.

Answer 1

First, prove it for the base case of$n=1$:
${11}^{(1+2)}+{12}^{(2+1)}$
= ${11}^3+{12}^3$
= $1331+1728$
= $133\left(10\right)+1+1728$
=$133\left(10\right)+1729$
=$133\left(10\right)+133\left(13\right)$
=$133\left(26\right)$
Assume it is true for a natural number $k$. Prove it is true for the number $k+1$:
True for $k$:
${11}^{(k+2)}+{12}^{(2k+1)}=133\left(m\right)$, where $m$ is an integer.
$\text{For}k+1:$
${11}^{k+1+2}+{12}^{2(k+1)+1}$
= ${11}^{k+2}\times 11+{12}^{2k+2+1}$
=${11}^{(}k+2)\times 11+{12}^{(}2k+1)\times {12}^2$
=$11\times {11}^{k+2}+(11+133)\times {12}^{2k+1}$
=$11({11}^{k+2}+{12}^{2k+1})+133\times {12}^{2k+1}$
=$11\left(133m\right)+133\times {12}^{2k+1}$
= $133(11m+{12}^{(2k+1)})$ hence it is divisible by $133$.