Solve $abc=2(a-2)(b-2)(c-2)$ where $a,b $ and $c$ are integers

Let $a \leq b \leq c$ be positive integer solutions of $abc = 2(a - 2)(b - 2)(c - 2)$. First note that

\[        (1 - 2/a)(1 - 2/b)(1 - 2/c) = 1/2.             (1)\]

Since $1 - 2/a \leq 1 - 2/b \leq 1 - 2/c$,  from (1) we conclude that $1 - 2/a \leq  \sqrt[3]{1/2}$, and so $a \leq 2/(1 - \sqrt[3]{1/2})$. So $a \leq 9$. 

Expanding $abc = 2[abc - 2(ab + bc + ca) + 4(a + b + c) - 8].$ Hence

\[abc - 4(ab + bc + ca) + 8(a + b + c) - 16 = 0.      (2)\]

Geometrically, it's clear a > 3. We consider the cases $a = 4, 5, \dots 9$ separately.

If $a = 4$. Then from (2) we have 
\[-16(b + c) + 8(b + c) + 16 = 0.\]
Therefore $b + c = 2$, contradicting $a \leq b \leq c$, and hence no solution.

If $a = 5$. Then
\[bc - 12(b + c) + 24 = 0  \Rightarrow (b - 12)(c - 12) = 120.\]
Hence
\[b - 12 = \{1, 2, 3, 4, 5, 6, 8, 10\}      and      c - 12 = \{120, 60, 40, 30, 24, 20, 15, 12\}.\]
Therefore
\[(b,c) = \{(13,132), (14,72), (15,52), (16,42), (17,36), (18,32), (20,27), (22,24)\}. \]

If $a = 6$. Then
\[2bc - 16(b + c) + 32 = 0 bc - 8(b + c) + 16 = 0    \Rightarrow     (b - 8)(c - 8) = 48.\]
Therefore
\[(b,c) = \{(9,56), (10,32), (11,24), (12,20), (14,16)\}.\]

If $a = 7$. Then
\[3bc - 20(b + c) + 40 = 0    \Rightarrow    9bc - 60(b + c) + 120 = 0    \Rightarrow    (3b - 20)(3c - 20) = 280.\]
Therefore
\[(b,c) = \{(7,100), (8,30), (9,20), (10,16)\}.\]

If $a = 8$. Then
\[4bc - 24(b + c) + 48 = 0    \Rightarrow   bc - 6(b + c) + 12 = 0     \Rightarrow   (b - 6)(c - 6) = 24.\]
Therefore
\[(b,c) = \{(8,18), (9,14), (10,12)\}\]
If $a = 9$. Then 
\[ 5bc - 28(b + c) + 56 = 0  \Rightarrow    25bc - 140(b + c) + 280 = 0    \Rightarrow    (5b - 28)(5c - 28) = 504. \] This has no solutions with $b \geq  9$.

The total number of solutions is 20.