Show that this expression is divisible by $133$ [more info in desc.]

Question

I'm sorry, but I just cannot figure this out to save my life. I need to show that the expression:

$$144*11^{n+1}+11*12^{2n-1}$$

is divisible by $133$ for all integers $n\geq 1$. How would I do this?

Answer 1

Let consider $\pmod{133}$

$$144\cdot11^{n+1}+11\cdot12^{2n-1} \equiv 11^{n+2}-12^{2n-2}\equiv 11^{n+2}-(12^2)^{n-1}\equiv11^{n+2}-11^{n-1}$$

$$\equiv 11^{n-1}\cdot(11^3-1)= 1330\cdot 11^{n-1}\equiv 0$$

Answer 2

if $n\geq 1$, then $$144*11^{n+1}+132*144^{n-1}=11^{n+2}-11^{n-1}=(1331-1)*11^{n-1}=0$$ where everything is considered modulo $133$.

Answer 3

$144=11$ mod $133$, $11*12^{2n-1}=11*12*({12^2})^{n-1}=132*(144)^{n-1}=132*11^{n-1}$, this implies that $144*11^{n+1}+11*12^{2n-1}=11^{n+2}+132*11^{n-1}$ mod $133$

$=11^{n-1}(11^3+132)=0$ mod $133$