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April 10th, 2017, 09:28 AM   #1
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(3×9^n+1)/(6 m -1) is not an integer

Show that $\dfrac{3\cdot 9^n +1}{6m -1}\not\in\mathbb{N}\;(\forall m,n\in\mathbb{N}^+).$
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April 11th, 2017, 08:02 AM   #2
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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
${\small\dfrac{3\cdot 9^n+1}{6m-1}}\not\in\mathbb{N}\;(\forall m,n\in\mathbb{N}^+)$
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