I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$.
I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them!
What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$.
Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$.
If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$.
I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.
- 7$\begingroup$ Umm, $\log 1=0$ so $\log 1 +\log 2+\log 3\neq \log 1 \log 2\log 3$. $\endgroup$Thomas Andrews– Thomas Andrews2013-12-19 16:37:44 +00:00Commented Dec 19, 2013 at 16:37
- 1$\begingroup$ Your equality with logs is not correct. RHS is 0 and LHS is $\log(2) + \log(3)$. $\endgroup$Jeremy Daniel– Jeremy Daniel2013-12-19 16:38:07 +00:00Commented Dec 19, 2013 at 16:38
- $\begingroup$ You can show that if $\frac{1}{x}+\frac1{y}+\frac 1{z}=1$ with $x,y,z$ natural, then either $x,y,z=3,3,3$ or $x,y,z=2,3,6$.Then which $a,b,c$ can we choose to get either of these solutions? $\endgroup$Thomas Andrews– Thomas Andrews2013-12-19 16:40:45 +00:00Commented Dec 19, 2013 at 16:40
- 3$\begingroup$ Your division, and size considerations, do the job. One of the terms must be $\frac{1}{2}$, meaning that one variable is $1$ and another is $2$. The end. $\endgroup$André Nicolas– André Nicolas2013-12-19 16:41:30 +00:00Commented Dec 19, 2013 at 16:41
- 1$\begingroup$ I think that the post I linked has the best answers (thus my vote). But it seems the mod has spoken. $\endgroup$Arctic Char– Arctic Char2021-12-18 21:13:26 +00:00Commented Dec 18, 2021 at 21:13
4 Answers
Without loss of generality $a \leq b \leq c$. Then $a+b+c \leq 3c$ and hence
$$abc=a+b+c \leq 3c$$
Thus, either $c =0$, in which case $a=b=c=0$, or
$$ab \leq 3 \,.$$
This leads to only four possibilities to check: $a=0$ or $(a,b)=(1,1)$ or $(a,b)=(1,2)$ or $(a,b)=(1,3)$.
- $\begingroup$ Using a similar argument, one can also solve $a+b+c+abc=0$ in natural numbers. $\endgroup$19021605– 190216052025-09-16 03:46:12 +00:00Commented Sep 16 at 3:46
- 1$\begingroup$ @19021605 In natural numbers all terms on the LHS in your equation are $\geq 0$. The only way to get $0$ is if all are $0$. $\endgroup$N. S.– N. S.2025-09-16 04:11:27 +00:00Commented Sep 16 at 4:11
- $\begingroup$ You're right. Maybe I meant in $\mathbb{Z}$ instead. $\endgroup$19021605– 190216052025-09-16 06:09:14 +00:00Commented Sep 16 at 6:09
If $a=0$ then you require $b+c=0$ and hence $b=c=0$.
Note that you can assume $a\leq b \leq c$. If $a, b, c \geq 2$ then $abc \geq 4c > c + b + a$. Hence at least one of $a,b,c$ is equal to $1$.
Wlog assume $a=1$, and look for solutions to $b+c+1 = bc$. If $b,c\geq 3$ then $bc \geq 3c > b + c + 1$, hence at least one of $b,c$ is less than $3$
Wlog assume $b=2$, and look for solutions to $c+3 = 2c$, which implies $c=3$.
So the only solutions are $(0,0,0)$ and $(1,2,3)$ and their permutations.
- 8$\begingroup$ From $b+c+1 = bc$, $(b-1)(c-1) = 2$. $\endgroup$ronno– ronno2013-12-19 17:02:28 +00:00Commented Dec 19, 2013 at 17:02
Here's a start of a full solution: The right side grows way faster than the left side, so it's unlikely that there are very many solutions. More formally, suppose that $a, b, c \ge 2$, and that $c$ is at least as large as $a, b$. Then we have
$$abc \ge 4c > c + c + c \ge c + b + a$$
so it's necessary that one of the numbers (which we'll call $a$) is $1$. So we can reduce the problem to studying
$$b + c = bc - 1$$
which has fewer variables.
- $\begingroup$ Thanks. I knew the solutions would contain inequalities, but couldn't find any except too general ones. $\endgroup$CODE– CODE2013-12-19 17:05:54 +00:00Commented Dec 19, 2013 at 17:05
leaving aside the solution in which $a=b=c=0$ order the numbers so $a \le b \le c$
as OP shows, $c|(a+b)$ so we have:
$$ \frac{a+b}{c} \le 2 $$
thus only the values $1$ and $2$ are possible for $\frac{a+b}c$. these give $a+b=c$ and $a+b=2c$ respectively. however the latter is only possible if all three numbers are equal, as $c$ cannot be the arithmetic mean of two smaller numbers.
if all three numbers are equal they must each be $\sqrt{3}$, not a natural number solution.
plugging $a+b=c$ back into the equation $a+b+c=abc$ gives $ab=2$ hence $a=1, b=2, c=3$
