1/ab+1/bc+1/ac=1 (a,b or c different 0)

So..? A more clear solution?

Using the symmetry $a \leq b \leq c$ and $a+b+c = abc \leq 3c $

Joker633 wrote: Using the symmetry $a \leq b \leq c$ and $a+b+c = abc \leq 3c $ That fails because integers are NOT natural numbers!

So... ?

liceeyes wrote: 1/ab+1/bc+1/ac=1 (a,b or c different 0) For positive integers Like he was doing. I'll use < (to mean less than or equal) _ yet to learn latex Let a< b < c ab<ac<bc ab = x ac = y bc = z 1 < 3/x x<3 x = 1, 2, 3 I think it's not so hard to solve from here for y and z and then for a, b , c

Ok thanks, but when you arrive to $-3\le ab\le 3$ how do you do to analize all the cases?

Let's study the cases when at least one of the numbers are $-1,$ $0$ or $1.$ If $c=-1 \Rightarrow~(a+1)(b+1)=2.$ So we get the solutions $(0,1,-1),~(-3,-2,-1)$ (and the permutations) If $c=0 \Rightarrow~ a+b=0.$ So we get some solutions: $(k,-k,0)$ (and the permutations) If $c=1 \Rightarrow~(a-1)(b-1)=2.$ So we get the solutions: $(3,2,1),~(0,-1,1)$ (and the permutations) Now let's suppose that the numbers aren't equal to $-1,~0$ or $1.$ We have $ \frac{1}{bc}+ \frac{1}{ac}+ \frac{1}{ab}=1.$ Note that $ \frac{1}{ac}+ \frac{1}{ab} \le 1.$ So $\frac{1}{bc} \ge 0 \Leftrightarrow bc>0.$ Similarly: $ac>0$ and $ab>0.$ Thus $ \text{sgn} (a)= \text{sgn} (b)= \text{sgn} (c).$ It's enough to solve the equation $\frac{1}{u}+ \frac{1}{v}+ \frac{1}{w}=1$ in $\mathbb{Z^+}.$ Let's suppose WLOG that $u \le v \le w.$ It follows that $u \le 3$ Checking all the cases, we will get the following solutions: $(u,v,w) \in \{(3,3,3),~(2,4,4),~(2,3,6)+ \text{permutations} \}.$ In order to get the rest of solutions, we have to solve the system $\begin{cases} bc=u \\ ac=v \\ ab=w \end{cases}$ in every case.

There are infinite possibilities...for example.... $a = -1, b = 0, c = 1$ $a + b + c = abc$ If we restrict the answer to positive integers then $a = 1 b = 2$ and $c = 3,$ may be the only possibility....

Is that correct, Little Kesha?

thanks ! I have finished