Geometric Log Arguments

$\begin{array}{l} {\log _6}a + {\log _6}b + {\log _6}c = {\log _6}abc = 6 \Leftrightarrow abc = {6^6} \\ \{ a,b,c\} {\text{ form an increasing geometric sequence, which means: }}{b^2} = ac \\ {\text{Therefore, }}{b^2} \times b = {6^6} \Rightarrow b = 36 \\ b - a{\text{ is a square of an integer, let the integer be }}k{\text{, then:}} \\ b - a = {k^2} \Rightarrow a = 36 - {k^2} = (6 - k)(6 + k) \\ ac = {b^2} = {6^4} \Leftrightarrow c = \displaystyle\frac{{{6^4}}}{{(6 - k)(6 + k)}} = \displaystyle\frac{{{2^4} \times {3^4}}}{{(6 - k)(6 + k)}} \\ c{\text{ is an integer, so }}{{\text{2}}^4}{\text{.}}{{\text{3}}^4}{\text{ must be divisible by }}(6 - k)(6 + k),\\ {\text{ i}}{\text{.e}}{\text{. }}(6 - k)(6 + k) \in {\text{ divisors of }}{{\text{6}}^4}.{\text{ }} \\ {\text{Obviously, }}k = 1,4,5{\text{ don't satisfy this condition.}}\\ {\text{For }}k = 2,(6 - k)(6 + k) = {2^2} \times {2^3} = {2^5},\\ {\text{which leaves a remaining 2 in the denominator}}{\text{.}} \\ {\text{So, only }}k = 3{\text{ satisfies the equation}}{\text{.}} \\ a = 36 - {3^2} = 27 \\ c = \displaystyle\frac{{{b^2}}}{a} = 48 \\ a + b + c = 27 + 36 + 48 = \boxed{111}. \\ \end{array}$

Let's write:

$a=\dfrac{k}{r}$

$b=k$

$c=kr$

$abc=k^3$

$\log_{6}{a}+\log_{6}{b}+\log_{6}^{c}=\log_{6}{abc}=\log_{6}{k^3}$

$\log_{6}{k^3}=6 \Rightarrow k=36$

So, $b=36$

Also, $36-\dfrac{36}{r}=m^2, m \in \mathbb{Z^{+}}$

With some trial and error, you will find that $r=\dfrac{4}{3}$ satisfies the above equation.

With that you get:

$a=27,b=36,c=48$

$a+b+c=111$

Q.E.D