# 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:

 $\ds a b c$ $\ge$ $\ds 16 c$ as $4 \le a \le b$ $\ds$ $\ge$ $\ds 4 \paren {a + b + c + c}$ as $a \le b \le c$ $\ds$ $>$ $\ds 4 \paren {a + b + c}$ as $c > 0$

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:

$\tuple {0, 0, 0}$

$\Box$

For $a \ge 1$, note that:

 $\ds 4 \paren {a + b + c}$ $=$ $\ds a b c$ $\ds \leadstoandfrom \ \$ $\ds 4 \paren {a + b}$ $=$ $\ds \paren {a b - 4} c$ $\ds \leadstoandfrom \ \$ $\ds c$ $=$ $\ds \frac {4 \paren {a + b} } {a b - 4}$

In the second step, left hand side 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:

 $\ds b c$ $\ge$ $\ds 9 c$ $\ds$ $\ge$ $\ds 4 \paren {b + c} + c$ as $b \le c$ $\ds$ $>$ $\ds 4 \paren {1 + b + c}$ as $c \ge b > 4$

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.

$\Box$

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

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

Suppose $b \ge 5$.

Then:

 $\ds 2 b c$ $\ge$ $\ds 10 c$ $\ds$ $\ge$ $\ds 4 \paren {b + c} + 2 c$ as $b \le c$ $\ds$ $>$ $\ds 4 \paren {2 + b + c}$ as $c \ge b > 4$

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.

$\Box$

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

Suppose $b \ge 4$.

Then:

 $\ds 3 b c$ $\ge$ $\ds 12 c$ $\ds$ $\ge$ $\ds 4 \paren {b + c} + 4 c$ as $b \le c$ $\ds$ $>$ $\ds 4 \paren {3 + b + c}$ as $c \ge b > 1$

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.

$\Box$

We have considered all possible values of $a$.

Hence the result.

$\blacksquare$