Fractional Diophantine Equation

Without loss of generality, assume that $a\leq b\leq c$ . Thus, we have $\begin{aligned} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{24}{abc}=1\leq \frac{3}{a}+\frac{24}{a^3}, \end{aligned}$ which gives $a^3-3a^2-24\leq 0$ . Thus, we have $a\leq 4$ and because $a\neq 1$ , we have $a\in\{2, 3, 4\}$ . We then do the casework:

Case 1: If $a = 2$ , then the given Diophantine equation becomes $\begin{aligned} \frac{1}{b}+\frac{1}{c}+\frac{12}{bc}=\frac{1}{2}\implies bc = 2b+2c+24, \end{aligned}$ which, using Simon's Favorite Factoring Trick (SFFT), transforms to $(b-2)(c-2)=28$ . Since $b\leq c$ means that $b-2\leq c-2$ , we deduce that $(b-2, c-2)\in\{(1, 28), (2, 14), (4, 7)\}$ , so we have $(a, b, c)\in\{(2, 3, 30), (2, 4, 16), (2, 6, 9)\}$ .

Case 2: If $a = 3$ , then the given Diophantine equation becomes $\begin{aligned} \frac{1}{b}+\frac{1}{c}+\frac{8}{bc}=\frac{2}{3}\implies 2bc = 3b+3c+24. \end{aligned}$ By SFFT, we have $(2b-3)(2c-3)=57$ . Since $b\leq c$ means that $2b-3\leq 2c-3$ and $b\geq 3\implies 2b-3\geq 3$ , we deduce that $(2b-3, 2c-3)\in\{(3, 19)\}$ , so we have $(a, b, c)\in\{ (3, 3, 11)\}$ .

Case 3: If $a = 4$ , then the given Diophantine equation becomes $\begin{aligned} \frac{1}{b}+\frac{1}{c}+\frac{6}{bc}=\frac{3}{4}\implies 3bc = 4b+4c+24. \end{aligned}$ By SFFT, we have $(3b-4)(3c-4)=88$ . Since $b\leq c$ means that $3b-4\leq 3c-4$ and $b\geq 4\implies 3b-4\geq 8$ , we deduce that $(3b-4, 3c-4)\in\{(8, 11)\}$ , so we have $(a, b, c)\in\{ (4, 4, 5)\}$ .

Therefore, the solutions of the given Diophantine equation are $\begin{aligned} (a,b,c)\in\{(2,3,30), (2, 4,16), (2,6,9),(3,3,11),(4,4,5)\} \end{aligned}$ and the sum of all possible values of $abc$ is $\begin{aligned} 180+128+108+99+80=\boxed{595}. \end{aligned}$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{24}{abc}=1$

$ab+ac+bc+24=abc$

$a+b+c+23=abc-ab-ac-bc+a+b+c-1=(a-1)(b-1)(c-1)$

$A=(a-1),B=(b-1),C=(c-1), A\le B\le C$

$A+B+C+26=ABC.$ Casework:

$A=1,B=2\implies C=29;A=1,B=3\implies C=15;A=2,B=2\implies C=10$

$(AC-1)(BC-1)=C(C+26)+1$

$C\lt 10\ ; C=8\implies A=1,B=5;C=4\implies A=B=3$