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 June 16th, 2018, 02:20 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, 08:11 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, 08:31 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,390 Thanks: 2015 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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