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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. Tags expression, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post restin84 Algebra 2 March 30th, 2012 05:06 PM norlyda Complex Analysis 9 January 7th, 2012 08:38 AM extatic Computer Science 0 April 28th, 2011 08:15 PM woodman5k Algebra 2 October 10th, 2007 04:53 PM extatic Applied Math 0 December 31st, 1969 04:00 PM

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