If $a-b,\, a+b\in T$ for all $a, b\in T$ then $T=\{mx:x\in \mathbb{Z}\}$ How to prove that if $T$ is a nonempty set of integer number so that $a-b,\, a+b\in T$ for all $a, b\in T$ then $T=\{mx:x\in \mathbb{Z}\}$ for some $m\in\mathbb{Z}$.

HINT: First show that if $A$ has just one element, then $A=\{0\}$, and we can take $m=0$. Now assume that $A$ has at least two elements.


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*Show that $A$ contains a positive integer.  

*Let $m=\min\{a\in A:a>0\}$, and show that $A\supseteq\{mx:m\in\Bbb Z\}$. 

*Finally, get a contradiction if there is an $n\in A\setminus\{mx:x\in\Bbb Z\}$ and conclude that $A=\{mx:m\in\Bbb Z\}$.

