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suppose $A\subseteq \mathbb{N}$ and for any $a,b,c\in A$ with $a<b<c$ we have $$a(b+c)=ab+c$$ what are all $A$ with this property?!

here $\mathbb{N}=\{1,2,3,...\}$.

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We have $ac = c$, so with $a < b < c$ this forces $A$ to have at most 3 elements and $1 \in A$ (or it can have less that three elements, and then the premise is false). –  dtldarek Feb 11 '13 at 22:43

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up vote 3 down vote accepted

Apply the distributive property, subtract the common term, and you get $ac=c$, so $a=1$. Any three element subset that includes $1$ will force this. Any set with less than three elements will also meet the requirement as there is no $a,b,c$.

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What about $A = \varnothing$? –  dtldarek Feb 11 '13 at 22:46

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