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 April 10th, 2017, 09:28 AM #1 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 (3×9^n+1)/(6 m -1) is not an integer Show that $\dfrac{3\cdot 9^n +1}{6m -1}\not\in\mathbb{N}\;(\forall m,n\in\mathbb{N}^+).$ April 11th, 2017, 08:02 AM #2 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 Some background The original problem is to show $\big(m,n,q={\small\dfrac{(m+3)^n+1}{3m}}\in \mathbb{N}^+ \big) \implies 2\nmid q$ Clearly (1) $3\mid (m+3)^n+1\iff 3\mid m^n+1\iff (2\nmid n)\wedge (m\equiv_3 -1)$ (2) $m\mid (m+3)^n+1\iff m\mid 3^n+1$ (3) $2\mid (m+3)^n+1\iff 2\mid m$ What we have found, based above is that (a) $(n=1)\wedge q\in\mathbb{N}^+ \implies q = 1$ (b) $(0\le k\in\mathbb{Z},\,k\ne 2)\wedge (m=2^k(2d-1),\,d,\,q\in\mathbb{N}^+) \implies 2\nmid q$ So we need to check the case where $m=4(2d-1),\,d\in\mathbb{N}^+$. By (1), $2\nmid n$ and so in this case $(m+3)^n+1=(8d-1)^n+1\equiv (-1)^n+1\equiv 0\pmod{8}$ So $q$ must not be an integer or we disprove the original claim. But $m=4(2b-1)\equiv 2\pmod{3}\iff m = 4(6d-1)$ and $3^{2p+1}=3\cdot 9^p$. So we need to show ${\small\dfrac{3\cdot 9^n+1}{6m-1}}\not\in\mathbb{N}\;(\forall m,n\in\mathbb{N}^+)$ Tags 1 or 6, 3×9n, integer 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 LaryIchigo Calculus 4 March 14th, 2014 09:44 PM BenFRayfield Number Theory 11 July 14th, 2013 10:56 AM harrypham Number Theory 1 July 30th, 2012 11:42 PM proglote Number Theory 8 December 25th, 2011 03:20 PM tinynerdi Number Theory 4 August 8th, 2010 07:28 PM

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