Find all finite set S such that any $a, b, c \in S$, $ab + bc + ca \in S$

Question

Find all finite set of real numbers S such that: $ab + bc + ca \in S$ with any distinct $a, b, c \in S$

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I just can solve the problem when $\exists$ at least $3$ elements $\in S\ge 1$

In another case, I got stuck.

Answer 1

Any set of cardinality $< 3$ is (trivially) a solution.

There are three-element solutions $\{a, b, c\}$ with $b = \dfrac{a(1-c)}{a+c}$ for any $a$ and $c$ such that $a \ne -c$ and $a, \dfrac{a(1-c)}{a+c}, c$ are distinct.

There are four-element solutions in complex numbers, e.g.

$$ \eqalign{a&=\dfrac{1}{3}+\dfrac{\sqrt {13}}{6}+\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr b & =\dfrac{1}{3}+\dfrac{ \sqrt {13}}{6}-\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr c &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}-\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}+\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr d &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}+\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}-\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr}$$

Answer 2

Since tge condition is about distinct elements, all $S$ with less than three elements will do.

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Let's check three-element sets $S=\{a,b,c\}$. Then $ab+ac+bc\ne0$ for if it were, say $a=0$, then $0=ab+ac+bc=bc$ and oe of $b,c$ equals $0$ contrary to the elements being distinct. So $0\notin S$ and wlog. $ab+ac+bc=a$. These sets can be parametrized as follows: Pick distinct $x,y\in\mathbb R\setminus\{2\}$ with $xy\ne1$. Then there is a unique choice of $a$ such that letting $b=a(x-1)$, $c=a(y-1)$ we have $a=ab+ac+bc$, namely $a=\frac{1}{(x-1)+(y-1)+(x-1)(y-1)}=\frac1{xy-1}$. (This explains the conditins on $x,y$ in hindsight: we need $x\ne y$ to have $b\ne c$, $x,y\ne2$ to have $b,c\ne a$, and $xy\ne1$ to find a solution)

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What about four-element sets $S=\{a,b,c,d\}$? Since $ab+ac+bc = (ab+ac+ad+bc+bd+cd)-(a+b+c+d)d+d^2=d^2+\alpha d+\beta$, where $f(X)=X^4+\alpha X^3+\beta X^2+\gamma X+\delta$ is the monic polynomial having the elements of $S$ as roots, we see that whenever $f(x)=0$ then $yx^2+\gamma x+\delta=0$ for some (not necessarily distinct) root $y$ of $f$. That is:

• If $\delta\ne0$ then $x\mapsto \frac{\gamma x+\delta}{x^2}$ is a map $S\to S$; if $\gamma=0$, the map goes $x\mapsto \frac\delta{x^2}\mapsto \frac{ x^4}{\delta}\mapsto \frac\delta{x^8}\mapsto\ldots$, which produces a divergent subsequence unless $x=\pm 1$. Since $S\setminus\{-1,1\}$ is nonempty we arrive at a contradiction and conclude $\gamma\ne 0$. I guess a lot of restrictions can be fround from this until one finds that no solution exists.
• if $\delta=0$ then $x\mapsto \frac \gamma x$ is a permutation of the nonzero roots (note that $\gamma\ne 0$ because we cannot have a double root). This permutation is an involution, hence $S$ consists of an even number of (i.e. two) elements of $\{-\sqrt \gamma,0,\sqrt \gamma\}$ and one pair of numbers with product $\gamma$. In other words, $\gamma=u^2>0$ and $S=\{0,u, t,\frac {u^2}t\}$ for some $t$. From $0,t,\frac{u^2}t$ we see $u^2 \in S$, which leaves the options $\{0,1,t,\frac1t\}$ and $\{0,u, 1,u^2\}$. The first requires $t+1+\frac1t\in S$, which is only possible if $t=-1$ (which is not allowed). The second requires $u^3\in S$, which is impossible.

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For $|S|=n>4$ I guess that the number $n\choose 3$ of triples is so much bigger than $n$ that the conditions cannot all be fulfilled.