<s>[INST] $\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$ $\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$
For which $n$ is this expression true. $n$ and $p$ are integers. $i$ and $j$ are positive integers or zero.
 [/INST] $$\frac{p}{2^i5^j} = \frac{6}{2n-1} - \frac{1}{n} = \frac{4n+1}{n(2n-1)}.$$
$n=1, 2$ are clearly solutions, and $3$ is not a solutions. Henceforth let $n\geq 4$.
Since $\gcd(n, 4n+1) = 1$, hence we must have $n \mid 2^i 5^j$. Since $\gcd(n, 2n-1) = 1$, they do not share any common factors. Note that $ \gcd( 2n-1, 4n+1) = \gcd(2n-1, 3 )$, hence is either 3 or 1. We consider cases as follows
Case 1: $\gcd(2n-1, 4n+1) = 1$.
Then one of $n$ and $2n-1$ is a power of 2 and the other is a power of 5. Since $2n-1$ is odd and greater than 7, it is the power of 5, and we have $ n = 2^i, 2n-1 = 5^j$. This gives us $$2^{i+1} - 1 = 5^j.$$ 
Consider mod 4. Since $n \geq 4$, hence $i\geq 2$, and so the LHS is $3 \pmod 4$, and the RHS is 1 $\pmod 4$,which is a contradiction. Thus, there are no solutions.
Case 2: $\gcd(2n-1, 4n+1) = 3$.
Since $2n-1$ is odd and greater than 7, we must have $n = 2^i$ and $ 2n-1 = 3 \times 5^j$. It is easy to check that $j=0, 1$ lead to the solutions $n=2, 8$. Henceforth, let $j \geq 2$.
Since $3 \mid 2^{i+1} - 1$, this tells us that $i$ is odd. Let $i = 2k + 1 $. This gives us
$$ 4^{k+1} -1 = 3 \times 5^j.$$
Dividing by $4 - 1 = 3$, we get $ 4^{k} + 4^{k-1} + \ldots + 1 = 5^j$. Since $j \geq 2$, we work mod 5 to get that $k+1$ must be even. Let $k = 2l + 1$. then dividing throughout by $4+1 = 5$, we get that
$$ 16^l + 16^{l-1} + \ldots + 1 = 5^{j-1}$$
Since $j-1 \geq 1 $, we work mod 5 to get that $l+1$ must be a multiple of 5. But this implies that $ 1 + 16 + 16^2 + 16^3 + 16^4 \mid 5^{j-1}$, which is clearly not possible since 69905 is not a power of 5.
Hence, the only solutions in this case are $n=2, 8$.
In conclusion, the only solutions are $n=1, 2, 8$.

I believe Case 2 has a simpler approach. I just bashed my way through it.
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