Triple with Sum and Product Equal

Theorem

For $a, b, c \in \Z$, $a \le b \le c$, the solutions to the equation:

$a + b + c = a b c$

are:

$\tuple {1, 2, 3}$

$\tuple {-3, -2, -1}$

and the trivial solution set:

$\set {\tuple {-z, 0, z}: z \in \N}$

Proof

Suppose one of $a, b, c$ is zero.

Then $a b c = 0 = a + b + c$.

The remaining two numbers sum to $0$, giving the solution set:

$\set {\tuple {-z, 0, z}: z \in \N}$

$\Box$

Suppose $a < 0$ and $0 < b \le c$.

Then $a b c \le a < a + b + c$.

Hence equality never happens.

Similarly, for $a \le b < 0$ and $c > 0$:

$a b c \ge c > a + b + c$

Hence equality never happens.

$\Box$

Now it remains the case $0 < a \le b \le c$.

Suppose $a \ge 2$.

Then:

\(\ds a b c\)

\(\ge\)

\(\ds 4 c\)

\(\ds \)

\(\ge\)

\(\ds a + b + c + c\)

\(\ds \)

\(>\)

\(\ds a + b + c\)

hence $a = 1$.

Suppose $b \ge 3$.

Then:

\(\ds b c\)

\(\ge\)

\(\ds c + c + c\)

\(\ds \)

\(\ge\)

\(\ds b + c + c\)

\(\ds \)

\(>\)

\(\ds b + c\)

hence $b = 1$ or $b = 2$.

For $b = 1$:

$c = 1 + 1 + c$

which is a contradiction.

For $b = 2$:

$2 c = 1 + 2 + c$

giving $c = 3$.

Therefore $\tuple {1, 2, 3}$ is the only solution for strictly positive integer values of $a, b, c$.

Similarly, for $a \le b \le c < 0$, we have $0 < -c \le -b \le -a$.

Also:

\(\ds \paren {-c} + \paren {-b} + \paren {-a}\)

\(=\)

\(\ds -\paren {a + b + c}\)

\(\ds \)

\(=\)

\(\ds -a b c\)

\(\ds \)

\(=\)

\(\ds \paren {-c} \paren {-b} \paren {-a}\)

and we have already show that $-c = 1, -b = 2, -a = 3$.

Therefore we have $\tuple {a, b, c} = \tuple {-3, -2, -1}$.

$\Box$

We have considered all possible signs of $a, b, c$.

Hence the result.

$\blacksquare$