Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$ Assume I have a function $f(n) = \frac{4n+1}{n(2n-1)}$ with $n \in \mathbb{N} \setminus \left\{ 0 \right\}$. The objective is to find all $n$ for which $f(n)$ has a proper decimal fraction. I know that any given fractiononly has a proper decimal fraction whenever the denominator only has the prime factors $2$ and $5$.
I wrote a script in Mathematica that calculated that for $n < 100000$ this only applies to $n_1 = 2 \lor n_2 = 8$. However this doesn't give me any explanation as to why this is the case. Assuming the fraction is already completely reduced: For $n(2n-1)$ to have the prime factors $2$ and $5$ only, $2n-1$ needs to have these prime factors as well. From this and the assumption I was able to figure that $n = \frac{1}{2} \left(5^i + 1 \right)$, whose last digit always is a $3$, which in turn means it is uneven. Also $n$ needs to have the prime factors $2$ and $5$ only, but as it needs to be uneven as shown before it can only have the prime factors $5$, which would however mean the last digit would have to be a $5$. I conclude that for every $n$ for which $f(n)$ has a proper decimal fraction, $f(n)$ cannot be already completely reduced, which means there are additional prime factors in the denominator that are also in the numerator. This coincides with the results the Mathematica script gave me, but I have no idea how to go about finding these common prime factors and possibly eliminating them, etc. 
I would greatly appreciate any help here!
Regards,
bk1ng
 A: $4n+1$ is relatively prime to $n$, and $\gcd(4n+1, 2n-1)$ is either $1$ or $3$.
So $n$ must be a power of $2$ times a power of $5$, and $2n-1$ is either that or $3$ times that.  If $n$ is divisible by $2$, $2n-1$ is not; if $n$ is divisible by $5$, $2n-1$ is not.  So you have four possibilities to consider:
1) $n = 2^p$, $2n-1 = 3 \times 5^q$, thus $2^{p+1} - 3 \times 5^q = 1$
2) $n = 2^p$, $2n-1 = 5^q$, thus $2^{p+1} - 5^q = 1$.
3) $n = 5^p$, $2n-1 = 3 \times 2^q$, thus $2 \times 5^p - 3 \times 2^q = 1$
4) $n = 5^p$, $2n-1 = 2^q$, thus $2 \times 5^p - 2^q = 1$
(2) is ruled out by Catalan's conjecture (Mihailescu's theorem).
(4) is ruled out by congruence mod $2$.
I think the others can be looked at using Størmer's theorem.
A: Note that $\gcd(4n+1,n)=1$ and $\gcd(4n+1,2n-1)=\gcd(3,2n-1)\in\{1,3\}$.
Thus $n$ must be of the form $n=2^a5^b$ and the odd number $2n-1$ must be of the form $2n-1=5^c$ or $2n-1=3\cdot 5^c$ with $a,b,c\in\mathbb N_0$. So from $1=2\cdot n-(2n-1)$ we find
$$\tag1 1 = 2^{a+1}5^b-5^c\qquad \text{or}\qquad 1=2^{a+1}5^b-3\cdot 5^c.$$In both cases $(b\ge1\land c\ge1)$ leads to a contradiction as the right hansd side then is a multiple of $5$, hence $b=0$ or $c=0$. 
The case $c=0$ is easy: $1=2^{a+1}5^b-1$ leads to $a=b=0$, $\color{red}{n=1}$; and $1=2^{a+1}5^b-3$ leads to $a=1, b=0$, i.e. $\color{red}{n=2}$. Remains the case $b=0$ and $c>0$, i.e. 
$$\tag2 1=2^{a+1}-(1\text{ or }3)\cdot5^c.$$
Since $c>0$, by taking remainders modulo $5$ we find in both cases
$$\tag3 1\equiv 2^{a+1}\pmod 5.$$
This implies that $a+1$ is a multiple of $4$, so $2^{a+1}=16^d$ for some $d$.
Rearranging $(2)$ we find
$$\tag4(1\text{ or }3)\cdot 5^c = 16^d-1=(4^d-1)(4^d+1).$$
At most one of the factor $4^d\pm1$ is a multiple of $5$, hence the other must be $\le3$. The only possibility is $d=1$, which leads to $a=3$, i.e. $\color{red}{n=8}$.
