Question: Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$.

Prove that $a = b$, then prove that $a$ is either $1$ or $2$.

I was thinking using the theorem:

> 
> If $n|a$ and $n|b$,then $n|(ax+by)$ for any $x,y\in \mathbb Z$.
> 
> 
> 

Then I can derive that $ab$ divides $a$ and $b$. I could not go further. However, this doesn't seem like can help me solve the question I had.

Answer: 
Your supposition is mistaken; $\wp(X\times X)$ is not the same as $\wp(X)\times\wp(X)$.

If you want actual equality as sets, then, as Rahul as mentioned, the elements of $\wp(X\times X)$ are sets of pairs of elements of $X$, while the elements of $\wp(X)\times\wp(X)$ are pairs of sets of elements of $X$.

If you just want them to be naturally in bijection with one another... well, that doesn't work either. In general, $\wp(X)\times\wp(Y)$ is naturally in bijection with $\wp(X\amalg Y)$ (here $\amalg$ is disjoint union), which is rather different.

Other people have already pointed out how to show that $X\times X$ is discrete, so I'll skip that...
Let (a,b) be any point in X×X.  

Then {a}, {b} are open within X.  

Thus {a}×{b} = {(a,b)} is open within in X×X.  

Consequently, since every singleton in X×X is open, X×X is discrete.