Prove $133$ divides $11^{n+2}+12^{2n+1}$

Question

Prove by the Principle of Recursion that for any integer $n\geq 0$, $$11^{n+2}+12^{2n+1}$$ is divisible by 133. Hint: You should consider a scheme like the one used to solve a pair of linear equations in two unknowns. In particular, think about how you eliminate one variable from such a system of equations.

Wow, despite the hint, I really feel lost with this one. The steps to proof by recursion can be seen here: https://pastebin.com/K1DKedAc

I honestly don't have very much practice with this, so any help would be greatly appreciated. Hopefully you can at least get me started. Thank you.

Answer 1

Hint: $$11^{n+3}+12^{2n+3}=11(11^{n+2}+12^{2n+1})+12^{2n+1}(12^2-11).$$

Answer 2

for $n=0$ is all clear, we denote by $$T_n=11^{n+2}+12^{2n+1}$$ and $$133|T_n$$now we assume that $$133|T_n$$ we have to Show that $$133|T_{n+1}$$ with $$T_{n+1}=11^{n+3}+12^{2n+3}$$ and we get $$T_{n+1}-T_n=(11^{n+2}+12^{2n+1}\cdot 10+133\cdot 12^{2n+1}$$ so we have $$T_{n+1}=(11^{n+2}+12^{2n+1})\cdot 10+133\cdot 12^{2n+1}+T_n$$ and we get $$133|T_{n+1}$$

Answer 3

Not a proof by recurrence...

$$11^{n+2}+12^{2n+1} =0 \mod 133$$ $$11^{n+2}+(12^2)^n12 =0 \mod 133$$ $$11^{n+2}+(144)^n12 =0 \mod 133$$ $$11^{n+2}+(11)^n12 =0 \mod 133$$ $$11^{n}(11^2+12) =0 \mod 133$$ $$11^{n}133 =0 \mod 133$$

Answer 4

Use the Chinese remainder theorem to have smaller moduli: $$11^{n+2}+12^{2n+1}\equiv 0\mod133=7\cdot 19\iff\begin{cases}11^{n+2}+12^{2n+1}\equiv 0\mod7,\\11^{n+2}+12^{2n+1}\equiv 0\mod197.\end{cases}$$

• Modulo $7$ this amounts to $4^{n+2}-2^{2n+1}=2^{2n+4}-2^{2n+1}=2^{2n+1}(2^3-1)\equiv2^{2n+1}(1-1)=0$.
• Modulo $19$, observe that $11^2\equiv 7$ and $12^2\equiv (-7)^2\equiv 11$. Now $$11^{n+2}+12^{2n+1}\equiv 11^{n+2}+11^n\cdot 12=11^n(11^2+12)\equiv 11^n(7+12)\equiv 0.$$

Answer 5

Hint: $121 \cdot 11^n+12 \cdot 144^n$ $= (133-12)\cdot 11^n + 12\cdot 144^n$ $= \color{red}{133} \cdot 11^n+ 12\cdot (\color{red}{144-11})(\ldots)$