As you can see from the title, I am trying to find all integer solutions $(a,b,c)$ to $$(a+b+c) \ \lvert\ (ab+bc+ca) \ \lvert\ abc$$ (that is, $a+b+c$ divides $ab+bc+ca$, and $ab+bc+ca$ divides $abc$). Unfortunately, I could not find anything on this problem (although I find it hard to believe that nobody though of this before).

What I've found so far

I have looked at the simpler case: $(a+b) \ \lvert\ ab$. I was able to solve this, and all solutions are $$(a,b)=(\alpha(\alpha+\beta)\gamma,\beta(\alpha+\beta)\gamma)$$ with $\alpha,\beta,\gamma\in\mathbb{Z}$.

I was also able to reduce the given problem to only one division. If we are able to solve $$(a_0b_0+b_0c_0+c_0a_0) \ \lvert\ (a_0+b_0+c_0)a_0b_0c_0 $$ with $\gcd(a_0,b_0,c_0)=1$, then we know that \begin{align} a&=a_0(a_0+b_0+c_0)\cdot k\\ b&=b_0(a_0+b_0+c_0)\cdot k\\ c&=c_0(a_0+b_0+c_0)\cdot k\\ \end{align} For $k\in\mathbb{Z}$ are all solutions to the original problem. However, I was not able to solve this. I have computed a few solutions to the last (and the corresponding solutions for the original problem) but was not able to find a pattern. Any progress on the problem is welcome!

  • 3
    $\begingroup$ These are coefficients of a polynomial $p(x)=(a-x)(b-x)(c-x)$. Maybe this helps... it seems like it should matter. $\endgroup$ – orion Jan 19 '16 at 10:16

$a,b,c$ are roots of $X^3-(a+b+c)X^2+(ab+ac+bc)X=abc$.

Putting $$\begin{cases}a+b+c=A\\ab+ac+bc=mA\\abc=n(ab+ac+bc)\end{cases}\qquad (*)$$ one gets $$X^3-AX^2+mAX=mnA$$ it follows $$\begin{cases}a^3-Aa^2+mAa=mnA\\b^3-Ab^2+mAb=mnA\\c^3-Ac^2+mAc=mnA\end{cases}$$ Hence $a^3\equiv b^3\equiv c^3\pmod A$ so that $a^3+b^3+c^3=kA$.

We take a look to the equation $$X^3+Y^3+Z^3=r$$ for cases in which there are solutions and we try to condition these to our problem.

Example.-For $r=s^3$. $$X^3+Y^3+Z^3= s^3 \Rightarrow (X,Y,Z)=(t,-t,s)$$ To fit with (*) we choose $t=2s$ which gives $(a,b,c)=(2s,-2s,s)$ so one has $$\begin{cases}a+b+c=2s-2s+s=s\\ab+ac+bc=-4s^2=(-4s)s\\ abc=-4s^3=(-4s^2)s\end{cases}$$ Thus $$a+b+c=s$$ $$\frac{ab+ac+bc}{a+b+c}=-4s$$ $$\frac{abc}{ab+ac+bc}=s$$ I will return to this problem…., if I may.

  • $\begingroup$ Nice answer. +1. $\endgroup$ – YoTengoUnLCD Jan 24 '16 at 20:35

Your original 2-variable problem can be written as "find all polynomials of the form $x^2+u x +uv$ with all integer zeroes". The conditions on what $u$ and $v$ must be come from the requirement that the zeroes are integers -- in this case, $-u \pm \sqrt{u^2-4uv}\in 2\mathbb{Z}$. This is satisfied when $u^2-4uv$ is a perfect square. In your case, $u=(\alpha+\beta)^2\gamma$ and $v=\alpha\beta\gamma$ satisfy this condition.

Your full problem can likewise be reduced to "find all polynomials of the form $x^3+ux^2+uvx+uvw$ with all integer zeroes". I'm not sure this is closer to a solution, but it somehow seems a better posed problem.

It seems to me that it should also be reducible to matrix algebra, but I don't quite see how.

  • $\begingroup$ How did you go from "...$u^2-4uv$ is a perfect square. ..." to "... $u=(\alpha+\beta)^2\gamma$ and $v=\alpha\beta\gamma$ ..."? $\endgroup$ – vrugtehagel Jan 19 '16 at 11:11
  • $\begingroup$ In this case, I just used your notation and verified it works. The $\gamma$ part can be done by observing simple scaling. The rest resembles a little bit the construction of all pythagorean triples (check wiki for the formulas) - it's completing to a perfect square. $\endgroup$ – orion Jan 19 '16 at 16:03

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