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June 16th, 2018, 03:20 AM  #1 
Newbie Joined: Nov 2016 From: Bulgaria Posts: 6 Thanks: 0  Proof of expression
Hello! How would one prove that $\displaystyle \forall n \in \mathbb{N} : 48^2 \mid 86^{n+1} + (48n  86)134^n$? I already checked that indeed $\displaystyle q = \frac{86^2 + (48*2  86)134^2}{48^2} \in \mathbb{Z}$ which is required for the divisibility theorem that states $\displaystyle \forall a, b \in \mathbb{Z} \Rightarrow \exists! \ q, r \in \mathbb{Z} : a = qb + r \land 0 \leq r < \lvert b \rvert $, but I seem to struggle with the second and third steps of the induction. Is this the correct approach? 
June 16th, 2018, 09:11 AM  #2 
Newbie Joined: Nov 2016 From: Bulgaria Posts: 6 Thanks: 0 
I suppose that since we know that $\displaystyle c \mid a \land c \mid (a + b) \Rightarrow c \mid b$ it would only make sense to subtract the "k" version from the "k + 1" version and prove that the result is a multiple of $\displaystyle 48^2$, but I didn't have any luck with it. Am I wrong or am I just not figuring out the numbers? 
June 16th, 2018, 09:31 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,004 Thanks: 1862 
I suggest expanding $86^{n+1} + (48n  86)(86 + 48)^n$ or $(134  48)^{n+1} + (48n  86)134^n$ by use of the binomial theorem. All the terms that don't obviously cancel have $48^2\!$ as a factor.


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