# All the integer solutions of a certain semi-algebraic system

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?

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.

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...

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

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

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

$$***$$

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

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

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

• Thanks for this answer! – aglearner Jul 12 '19 at 12:57