Triple with Product Quadruple the Sum

Theorem
Let $a, b, c \in \N$ such that $a \le b \le c$.

Then the solutions to:
 * $a b c = 4 \paren {a + b + c}$

are:
 * $\tuple {0, 0, 0}, \tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$

Proof
Suppose $a \ge 4$.

Then:

hence $0 \le a \le 3$.

For $a = 0$, we have $4 \paren {b + c} = 0$.

This forces $b = c = 0$, giving the trivial solution:
 * $\tuple {0, 0, 0}$

For $a \ge 1$, note that:

In the second step, is strictly positive.

Hence $a b > 4$ and the final step is justified.

For $a = 1$, we have $b c = 4 \paren {1 + b + c}$.

Since $a b > 4$, $b > 4$.

Suppose $b \ge 9$.

Then:

hence $b \le 8$.

We find the value of $c$ by substituting $a$ and $b$:
 * $b = 5: c = \dfrac {4 \paren {1 + 5} } {5 - 4} = 24$
 * $b = 6: c = \dfrac {4 \paren {1 + 6} } {6 - 4} = 14$
 * $b = 7: c = \dfrac {4 \paren {1 + 7} } {7 - 4} = \dfrac {32} 3$
 * $b = 8: c = \dfrac {4 \paren {1 + 8} } {8 - 4} = 9$

and we see that $\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}$ are valid solutions.

For $a = 2$, we have $2 b c = 4 \paren {2 + b + c}$.

Since $a b > 4$, $b > 2$.

Suppose $b \ge 5$.

Then:

hence $b \le 4$.

We find the value of $c$ by substituting $a$ and $b$:
 * $b = 3: c = \dfrac {4 \paren {2 + 3} } {2 \times 3 - 4} = 10$
 * $b = 4: c = \dfrac {4 \paren {2 + 4} } {2 \times 4 - 4} = 6$

and we see that $\tuple {2, 3, 10}, \tuple {2, 4, 6}$ are valid solutions.

For $a = 3$, we have $3 b c = 4 \paren {3 + b + c}$.

Suppose $b \ge 4$.

Then:

hence $b \le 3$.

Since $3 = a \le b$, this forces $b = 3$

We find the value of $c$ by substituting $a$ and $b$:
 * $c = \dfrac {4 \paren {3 + 3} } {3 \times 3 - 4} = \dfrac {24} 5$

and we see that it is not a valid solution.

We have considered all possible values of $a$.

Hence the result.