Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
Umm, $\log 1=0$ so $\log 1 +\log 2+\log 3\neq \log 1 \log 2\log 3$. –  Thomas Andrews Dec 19 '13 at 16:37
1  
Your equality with logs is not correct. RHS is 0 and LHS is $\log(2) + \log(3)$. –  Jeremy Daniel Dec 19 '13 at 16:38
    
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? –  Thomas Andrews Dec 19 '13 at 16:40
    
2  
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. –  André Nicolas Dec 19 '13 at 16:41

3 Answers 3

up vote 12 down vote accepted

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 $(a,b)=(1,3)$.

share|improve this answer

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.

share|improve this answer
6  
From $b+c+1 = bc$, $(b-1)(c-1) = 2$. –  ronno Dec 19 '13 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.

share|improve this answer
    
Thanks. I knew the solutions would contain inequalities, but couldn't find any except too general ones. –  CODE Dec 19 '13 at 17:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.