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I'm trying to evaluate the following fraction $\frac{3^a}{2^b-3}[(\frac{2^b}{3})^c - 1]$, but I'm getting stuck using gcd arguments or divisibility arguments. This is part of an ongoing research I'm helping a professor with within the number theory department.

Note: $a,b,c \in Z$

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  • $\begingroup$ Don't you have any conditions on $a,b,c$ ? I mean, take $a=0$, $b=1$ and $c=1$, you get $\frac{1}{3}$ if I'm not mistaken. $\endgroup$
    – M. Dus
    Commented Jan 25, 2018 at 8:31
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    $\begingroup$ It's equal to $3^{a-c}\frac{(2^b)^c-3^c}{2^b-3}$. Note that $\frac{(2^b)^c-3^c}{2^b-3}$ is always an integer (moreover, $\frac{a^k-b^k}{a-b}$ is always an integer for integer $a,b$), and the numerator is coprime to $3$, so this expression is an integer iff $3^{a-c}$ is an integer. Please use math.stackexchange.com for questions of this level. $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 25, 2018 at 10:52

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The formula can be rewritten as follows:

$$3^{a-1} \frac{(\frac{2^{b}}{3})^c - 1}{\frac{2^b}{3} - 1}$$

Using the general formula $\sum_{i=0}^{n} x^n = \frac{x^{n+1}-1}{x-1}$, this rewrites to

$$3^{a-1} \sum_{i=0}^{c-1} \left(\frac{2^b}{3}\right)^i$$

When writing the sum as a number in digits with respect to base 3, one can see that the $(c-1)$th digit after the decimal point is non-zero. Thus, $a-1$ has to be at least $c-1$. Hence, the condition that you are looking for is:

$$a\ge c.$$

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