Find all $M \in \mathbb{N}_{>0}$ s.t. $0 \equiv M^{n+1} + 12^{2n-1} \pmod{133}, \forall n \in \mathbb{N}_{>0}$ NB?: Holds for M = 11.
Not sure how to approach the above question?
 A: First note that $12^3 \equiv -1 \pmod{133}, 12^6 \equiv 1 \pmod{133}$, therefore $-12^{2n-1}$ cycles through $121, 1, 11, 121, 1, 11, \cdots, \pmod{133}$, and we must have $$M^2\equiv 121, M^3\equiv 1, M^4 \equiv 11 \pmod{133}$$
Then it follows immediately that $M\equiv M \cdot M^3 = M^4 \equiv 11 \pmod{133}$, and we can easily check that $M^{n+1}$ cycles through $121, 1, 11, 121, 1, 11, \cdots, \pmod{133}$.
Conclusion: $M=11+133(k-1), \forall k \in \mathbb N$.
A: 
how to approach the above question?

Here's a way.  In order for $M$ to qualify, it would have to work for $n=1$, so we'd have $M^2+12\equiv0\bmod133$.  That happens iff $12+M^2\equiv0\bmod7$ and $12+M^2\equiv0\bmod19$; equivalently, $M^2\equiv2\bmod7$ and $M^2\equiv7\bmod19$; equivalently, $M\equiv\pm3\bmod7$ and $M\equiv\pm8\bmod19$; equivalently, $M\equiv 11, 46, 87, $ or $122\bmod133$.  We have already narrowed down the possibilities for $M$ considerably.  Now check if any of those $M$ values works for $n=2$, and, if one does, see if you can prove it works for all $n\in \mathbb N_{>0}$.
P.S.  After writing the above, I thought of this alternative approach.
In order for $M$ to qualify, it would have to work for $n=\phi(133)=108$.    Then we'd have $M^{109}+12^{215}\equiv0\bmod133$, which is equivalent to $M\equiv-144^{108}12^{-1}\equiv-(-11)=11.$  Now you just have to show that, if $M\equiv11\bmod133$, then it works for all $n\in \mathbb N_{>0}$, and it does, because $11^{n+1}+12^{2n-1}=121\cdot11^{n-1}+12\cdot144^{n-1}\equiv133\cdot 11^{n-1}\equiv0\bmod133$.
