2015/046) Prove that for any natural number n, $11^{n + 2} + 12^{2n + 1}$ is divisible by 133

we have

$11^{n + 2} + 12^{2n + 1}$

= $11^2 * 11 ^n + 12 * 12^{2n}$

= $121 * 11^n + 12 * 144^n$

= $121 * 11^n + 12 * 11^n + 12 * 144^n – 12 * 11^n$

= $133 * 11^n + 12( 144^n – 11^n)$

the 1^st term is multiple of 133 and 2^nd term is divisible by 144-11 $( a^n-b^n)$ is divisible by a – b and hence the sum.