Prove that $|A|=2$ Let $A$ be a set of non-zero natural numbers, $|A| \geq 2$, which has the following property: $\forall a,b \in A, a>b$, we have $\frac{[a,b]}{a-b}\in A$  (where $[x,y]$ is the least common multiple of $x$ and $y$).
Prove that $A$ has exactly $2$ elements.
I started assuming $b=\min A$ and $a \in A-\{b\}$.
From $(a-b)|[a,b]$, it follows that $(a-b)|ab.$
Since $(a-b)|(a-b)$, we obtain that $(a-b)|(ab-b(a-b))$, so $(a-b)|b^2.$
We find out that $a-b \leq b^2$, hence $a \leq b+b^2$, which means $A$ is finite.
That's all I've managed to do.
 A: Let the two smallest numbers be $b$ and $a$, such that $b < a$. Let their gcd be $k$. So $a = a_1k$ and $b=b_1k$.
Now $a_1b_1k \over a_1-b_1$ is an integer but we know $(a_1,b_1)=1$ so $a_1-b_1 | k$ or $a_1-b_1 = 1$, where the latter case trivially implies $a_1-b_1 | k$ as well.
So let's write $k = m(a_1-b_1)$ so  $a_1b_1k \over a_1-b_1$$ = a_1b_1m$.
We have three numbers in the following order
$$b_1m(a_1-b_1) < a_1m(a_1-b_1) \leq a_1b_1m$$
Now take the third number and the first number,
We know $lcm(a_1b_1m,$ $ b_1m(a_1-b_1))$ must be a divisor of $a_1b_1m(a_1-b_1)$.
So $a_1b_1m - b_1m(a_1-b_1) = b_1^2m | a_1b_1m(a_1-b_1)$
Or $b_1|a_1(a_1-b_1)$ which means $b_1=1$
Look at the second and the third numbers
$a_1m(a_1-b_1) \leq a_1b_1m \implies a_1-1 \leq 1$ which means $a_1=2$.
So we have two smallest numbers $k$ and $2k$ in our set.
If there exists any $c>2k$ in the set, then 
$${ck\over c-k}\geq2k \implies c \geq 2c-2k \implies 2k \geq c$$
Or 
$${ck\over c-k} = k \implies c =c-k \implies k=0$$
Both of which are impossible.
