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April 10th, 2017, 10:28 AM  #1 
Senior Member Joined: Dec 2006 Posts: 166 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, 09:02 AM  #2 
Senior Member Joined: Dec 2006 Posts: 166 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) $(0\le k\in\mathbb{Z},\,k\ne 2)\wedge (m=2^k(2d1),\,d,\,q\in\mathbb{N}^+) \implies 2\nmid q$ So we need to check the case where $m=4(2d1),\,d\in\mathbb{N}^+$. By (1), $2\nmid n$ and so in this case $(m+3)^n+1=(8d1)^n+1\equiv (1)^n+1\equiv 0\pmod{8}$ So $q$ must not be an integer or we disprove the original claim. But $m=4(2b1)\equiv 2\pmod{3}\iff m = 4(6d1)$ and $3^{2p+1}=3\cdot 9^p$. So we need to show ${\small\dfrac{3\cdot 9^n+1}{6m1}}\not\in\mathbb{N}\;(\forall m,n\in\mathbb{N}^+)$ 

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1 or 6, 3×9n, integer 
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