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
On
The question originally left out the condition that $a$ and $b$ are integers.
Without that condition, the statement is not true.
Consider the counter-example:
$a=\frac 25$ and $b= \frac 27$
$\frac 1a + \frac 1b= \frac 52 + \frac 72 = 6$
Now that the question has been altered, this is no longer an answer, but I shall leave it here unless downvoted.
On
A variant (supposing $a,b$ are natural numbers):
Say $a\le b$. The hypothesis means $ab=a+b\le 2b$, whence $a\le 2$. Furthermore $b\mid a+b$, hence $a+b-b=a$, so that $a=b$.
On
If $\frac1a + \frac1b = \frac{a+b}{ab}$ is an integer, then $a$ and $b$ both divide $a+b$. This means that $a$ divides $(a+b)-a=b$ and that $b$ divides $(a+b)-b=a$. So $a$ and $b$ divide each other, whence they must be equal.
Now if $\frac1a + \frac1b = \frac2a$ is an integer, then $a$ must divide $2$. So $a=1$ or $a=2$.
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 $a=2$.