Number Theory – Prove $11^{n+1}+12^{2n-1}$ is Divisible by $133$

divisibilityelementary-number-theoryinduction

Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. Most likely, due to the late hour, I am missing something very trivial here.

Prove that for any integer $n$, the number $11^{n+1}+12^{2n-1}$ is divisible by $133$.

I have tried multiplying through by $12$ and rearranging, but like I said with meager results. I arrived at $11^{n+2}+12^{2n+1}$ which satisfies the induction hypothesis for LHS, but for the RHS I got stuck at $12 \times 133m-11^{n+1}-11^{n+3}$ or $12^2 \times 133m-12 \times11^{n+1}-11^{n+3}$ and several other combinations none of which would let me factor out $133$.

Best Answer

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

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