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We say that a set $A$ is "good" if $A\subseteq\mathbb{R}$, and for every positive integer $n$, if $a_1,\dots,a_n\in A$, and for every $1\leq i,j\leq n$ with $i\neq j$ we have $a_i\neq a_j$, then $$\sum_{k=1}^{n}a_k=\prod_{k=1}^{n}a_k.$$

I conjecture that there is no "good" set $A$ with $|A|>1$.

I considered this question after seeing this post. If we take $A=\{1,2,3\}$, then $1+2+3=1\cdot 2\cdot 3$, but, for example, we have $1+2\neq 1\cdot 2$. Clearly, if $|A|\in\{0,1\}$, then $A$ is "good." If $|A|>1$ and $0\in A$ then there is an $x$ with $x\in A$ and $x\neq 0$, so that $x+0=x\cdot 0\implies x=0$ which is a contradiction. In a similar manner we get a contradiction if $1\in A$. Thus, one approach is to show that if $A$ is "good" and $|A|>1$ then $0\in A$ or $1\in A$. If we drop the restriction that for every $1\leq i,j\leq n$ with $i\neq j$ we have $a_i\neq a_j$, then for every $a\in A$ we would get $a^2=2a$, which gives $a=0$ or $a=2$.

Is my conjecture true?

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    $\begingroup$ As written, there are no finite good sets, because you require your property to hold for every $n$, not just $n \leq |A|$. $\endgroup$
    – orlp
    Commented yesterday
  • $\begingroup$ Naïvely, your requirement boils down to $2^n - n - 1$ equations (one for each subset with more than one element) in $n$ unknowns. This leads to one equation in two unknowns when $n = 2$, but more equations than unknowns when $n > 2$. So we would expect one degree of freedom when $n = 2$ (see Servaes' answer) but we would expect the system to be overdetermined (i.e. no solutions) when $n > 2$. This is far from a rigorous proof, though. $\endgroup$ Commented yesterday
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    $\begingroup$ @orlp A set is 'good' if a clearly specified implication holds for every positive integer $n$: $$\text{If }a_1,\ldots,a_n\in A\text{ are pairwise distinct then }\sum_{i=1}^na_i=\prod_{i=1}^na_i.$$ The antecedent to this implication cannot hold if $n>|A|$, and so the implication is vacuously true for all $n>|A|$. $\endgroup$
    – Servaes
    Commented yesterday
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    $\begingroup$ @MichaelSeifert That is indeed far from a rigorous proof, and one would expect there to be some dependence between the equations due to symmetry, and so the system might not be as overdetermined. But somewhat surprisingly your simple heuristic comes out exactly right :) $\endgroup$
    – Servaes
    Commented yesterday

1 Answer 1

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Your conjecture is false: For any real number $a\notin\{0,1,2\}$ the two-element set $A:=\{a,\tfrac{a}{a-1}\}$ is good.


To see that these are the only nontrivial good sets, note that if $A$ is a good set with $|A|=2$ then $A=\{a,\tfrac{a}{a-1}\}$ for some real number $a$. If $A$ is a good set then for any subset $A'\subset A$, also $A'$ is a good subset. So for a good set $A$, all its $2$-element subsets are of the form $\{a,\tfrac{a}{a-1}\}$ for some real number $a$. Then of course $A=\{a,\tfrac{a}{a-1}\}$ for some real number $a$.

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    $\begingroup$ And this is the only good set with $a$ because, if $ab=a+b$ then $b=a/(a-1)$. $\endgroup$ Commented 2 days ago

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