Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$

See that $12^2 = 144 \equiv 11 \pmod{133}.$

So we have

$$11^{n+2} + 12^{2n+1} \equiv 121(11)^n +12(144)^n $$

$$\equiv 121(11)^n +12(11)^n \equiv 133(11)^n \equiv 0 \pmod{133}.$$

Notice $\ 133 = 11^2\!+\!11\!+1\ $ so we may apply

Lemma $ \bmod\ a^2\!+a+1\!:\,\ b := a^{n+2}\!+(a\!+\!1)^{2n+1}\!\equiv 0\,$ in any ring, $ $ for all $\,n\in\Bbb Z$

${\bf Proof}\,\ {\rm Note\ that}\ \ \color{#0a0}{a(a\!+\!1)\equiv -1}$ and $\,\color{#c00}{a^3\equiv 1}\ $ by $\,0\equiv (a\!-\!1)(a^2\!+a+1) = a^3\!-1,\,$ so

$\ a^{2n+1}b\, =\, a^{3n+3}\! + (\color{#0c0}{a(a\!+\!1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0\ $ so $\ b\equiv 0,\,$ by $\,a$ $\rm\color{#c00}{unit}$ so cancellable.

Or use $\ \color{#90f}{a\!+\!1}\equiv -a^2\,\Rightarrow\, (\color{#90f}{a\!+\!1})^{2n+1}\equiv -a^{4n+2}\equiv -(\color{#c00}{a^3})^n a^{n+2}\equiv -a^{n+2}$

Remark $ $ Both are essentially special cases of the method used here, which is a special case of the method of simpler multiples.

since $133=7*19$, first show that $$7\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$ then $$19\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$

Since $gcd(7,19)=1$, the result follows. $$ 11^{n+2}\equiv 11^{n}\cdot 11^2\equiv 2\cdot11^{n}\bmod 7$$ $$12^{2n+1}\equiv 5\cdot 144^{n}\equiv 5\cdot 11^{n}\bmod 7$$ So $$11^{n+1} + 12^{2n+1} \equiv 7\cdot11^n\equiv 0 \bmod 7$$ $$..........................................................................$$ $$11^{n+2}\equiv 121\cdot 11^n\equiv 7\cdot11^{n}\bmod 19$$ $$12^{2n+1}\equiv 12\cdot 144^{n}\equiv 12\cdot 11^{n}\bmod 19$$ So $$11^{n+1} + 12^{2n+1} \equiv 19\cdot11^n\equiv 0 \bmod 19$$