<s>[INST] If $a > 0$,$b>0$, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$ If $a$ and $b$ are positive integers, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$
-I played around with numbers and the conditions and it seems that it's common sense that if $a = b$ then $a = b = 1$ or $2$, For example:
If $a = b = 4$, then $\frac{1}{a} + \frac{1}{b}$ would not be an integer.
-Any help is appreciated
 [/INST] Suppose $a>b$, then $\frac{1}{a}+\frac{1}{b}<\frac{2}{b}$, and since $\frac{1}{a}+\frac{1}{b}$ is integer with $a>0, b>0$, $$\frac{2}{b}>\frac{1}{a}+\frac{1}{b}\geq1$$or, $2>b$, hence $2>b>0$, and $b=1$.  
Hence, $\frac{1}{a}+\frac{1}{b}=\frac{1}{a}+1$ is integer, and so $a=1$ (because $\frac{1}{a}$ is positive integer).
But this is a contradiction, since we supposed that $a>b$. Therefore $a \le b$.
Similarly, if we start by suposing that $a<b$ we find again that $a=b$ : a contradiction. So $a \ge b$.
Combining these two statements gives $a=b$ as required.
In addition, $\frac{1}{a}+\frac{1}{b}=\frac{2}{a}$ is integer, so, since $a>0$, we have $a=1$ or $a=2$.
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