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.
$\blacksquare$