How find all finite sets $ M$ such that $ |M|\ge 2$ and $ \frac {2a}{3} - b^2\in M$ for all $ a,b\in M$ How find all finite sets of real numbers $ M$ such that $ |M|\ge 2$ and $ \frac {2a}{3} - b^2\in M$ for all $ a,b\in M$?
 A: Assume that $M$ is a non-empty finite set with this property. Let $p$ be the smallest element of $M$ (it exists because $M$ is non-empty and finite). Then the number $2p/3-p^2\in M$, so we have
$$
p\le 2p/3-p^2.
$$
This inequality holds, iff $p\in[-1/3,0]$.
For all $b\in M$ we then have $2p/3-b^2\ge p$, so $b^2\le-p/3$. This gives us an
upper bound for $|b|\le \sqrt{-p/3}\le \sqrt{1/9}=1/3$. 
Next let us turn our attention to the element of $M$ that is closest to zero, i.e. one with the minimal absolute value. Call that element $\epsilon\in M$.
Given that $\epsilon$ is kinda small the number $\delta=2\epsilon/3-\epsilon^2\in M$ is small also. But by the choice of $\epsilon$ its absolute value cannot be smaller than $|\epsilon|$. 
If $\epsilon>0$ then $\delta<\epsilon$. By the minimality of $|\epsilon|$ we must then have $\delta\le-\epsilon$. This is equivalent to $5\epsilon/3\le\epsilon^2$, so $\epsilon\ge5/3$ contradicting our upper bound.
If $\epsilon=0$ then we can select $b=0$ together with all $a\in M$. This implies that $(2/3)^ka\in M$ for all $a\in M$ and $k\in\Bbb{N}$. That is an infinite sequence of distinct elements, so we can conclude that $0\notin M$.
If $\epsilon<0$, then also $\delta<0$. Therefore we must have $\delta\le\epsilon$. This is equivalent to $-\epsilon/3\le\epsilon^2$ and dividin by $\epsilon<0$ implies that $-1/3\ge\epsilon$. This leaves us only with the choice
$$
p=-\frac13=\epsilon.
$$
But now we are in trouble. All of the above implies that $M\subseteq\{-1/3,1/3\}$. But if $1/3\in M$, then so is $1/9=(2/3)\cdot(1/3)-(1/3)^2$ contradicting the choice of $\epsilon$.
Conclusion: Unless I misunderstood something, there are no such sets. The above considerations also imply that $\{0\}$ and $\{-1/3\}$ are the only singleton sets closed under that function.
