I would like to find all integer solutions of the following system:

$$a+b+c+ab+ac+bc=-2,$$ $$a,b,c\le a+b+c-1.$$

One solution is $2,2,-2$. Is it possible to describe all others?

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Here is a rough idea, the rest should not be too hard to fill in.

First, the given condition implies $a+b,b+c,c+a\geqslant 1$. Namely, $a+b+c$ is positive. Moreover, there is at least one negative number among, and not two. Suppose therefore $c<0$, and let $c=-d$ (I abuse the notation a bit here). Then, observe that, adding $1+abc$ to both sides, you get: $$ -1+abc=(a+1)(b+1)(c+1)\Leftrightarrow (a+1)(b+1)(d-1)=abd+1. $$ Now, I claim that you can recover a family of infinitely many solutions from here. Let $d=k+1$ for $k$ fixed. Then, $$ kab+ka+kb+k = (k+1)ab+1\Rightarrow ab-ka-kb+k^2 = k^2+k-1. $$ Namely, $$ (a-k)(b-k)=k^2+k-1. $$ Thus, the set of all solutions are of the following form: $$ \{(a,b,c):c=-k-1,a=k+r,b=\frac{k^2+k-1}{r}+k,r\mid k^2+k+1,k\in\mathbb{Z}^+\}. $$

**Edit:** Forgot to add. All permutations $(a,b,c)$ also work. The solution above is the set of all solution triples with $c$ being negative.

**Edit 2:** Typos fixed.

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To solve the Diophantine equation. Where $q$ is any given number...

$$XY+XZ+YZ+X+Y+Z=q$$

Solutions can be expressed through the following solutions to the Pell equation.

$$p^2-k(k+t)s^2=1$$

And then the decisions can be recorded...

$$X=(q+1)p^2-((q+3)k+(q+1)t)ps+k(2k+t)s^2$$

$$Y=p^2-((q+3)k+(q+2)t)ps+((q+2)k^2+(2q+3)kt+(q+1)t^2)s^2$$

$$Z=-p^2+((q+2)k+(q+1)t)ps$$

$$***$$

$$X=qp^2+(q+1)(k+t)ps+k^2s^2$$

$$Y=((q+1)k+qt)ps+(q+1)(k+t)^2s^2$$

$$Z=-((q+2)k+(q+1)t)ps+k(k+t)s^2$$

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