0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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...

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

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

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

$$***$$

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

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

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

$\endgroup$
  • $\begingroup$ Thanks for this answer! $\endgroup$ – aglearner Jul 12 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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