Where can I read the rules on the algabraic manipulation needed to finish this inductive proof that $133$ divides $11^{k+1}+12^{2k-1}$?

Question

Prove: 133 | $11^{k+1}$ + $12^{2k-1}$ , $\forall$ k > 0

Basis: Let k = 1
133 | $11^{2}$ + $12^{1}$ $\implies$ 133 | 133, True

Hypo: Assume true for k = n
133 | $11^{n+1}$ + $12^{2n-1}$

Need to Prove: k = (n+1)
133 | $11^{(n+1)+1}$ + $12^{2(n+1)-1}$
$\implies$ 133 | $11^{n+2}$ + $12^{2n+2-1}$
$\implies$ 133 | $11$($11^{n+1}$) + $12^2$($12^{2n-1}$)
???

I know that by the hypothesis, 133 divides what is in the parenthesis, and that $11+12^2$ is also divisible by 133, but how can I manipulate the equation to make it so? If you guys can refer me to where I can read about this, or explain to me how and why, I would appreciate it. Thank you.

Answer 1

$$11^{(n+1)+1} + 12^{2(n+1)-1} = 11 \cdot 11^{n+1} + 144 \cdot 12^{2n-1} = 11 \left(11^{n+1} + 12^{2n-1}\right) + 133 \cdot 12^{2n-1}$$

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• $11 \left(11^{n+1} + 12^{2n-1}\right)$ is divisible by 133 by the induction hypothesis.
• $133 \cdot 12^{2n-1}$ is clearly a multiple of $133$.

Answer 2

By the induction hypothesis, you know that $133|11^{n+1} + 12^{2n-1}$, meaning that for some integer $k$, you have that $$133k = 11^{n+1} + 12^{2n-1}$$ or equivalently $$12^{2n-1} = 133 - 11^{n+1}.$$ PLug this into you last expression you obtained and you'll find $$11(11^{n+1}) + 12^2(133 - 11^{n+1})$$ and after rewriting: $$133(12^2) + 11^{n+1}(11 - 12^2)$$ and $11 - 12^2 = 11-144 = -133$.