How do we deduce that a=b?

Question

If $a,b \in \mathbb{Z}^{+}$ and $\frac{1}{a}+\frac{1}{b} \in \mathbb{Z}$, I want to show that $a=b$. After that, I want to show that $a=1$ or $a=2$.

I have thought the following so far.

Since $\frac{1}{a}+\frac{1}{b} \in \mathbb{Z}$, it follows that $\frac{1}{a}+\frac{1}{b}=k, \text{ for some } k\in \mathbb{Z}^{+}$.

Then $b+a=kab$.

Suppose that $a \neq b$. Then $\exists$ some prime $p$ such that $p \mid a$ and some prime $q \neq p$ such that $q \mid b$.

Then $p \mid b+a$ and $p \mid a$. Thus $p \mid b$. So we get that $p \cdot q \mid b$.

In the same way, we get that $p \cdot q \mid a$.

Therefore, we get that $p \cdot q \mid a \cdot b$.

Can we use the above in order to deduce that $a=b$ ?

Answer 1

It is much simpler to use inequalities. We have $\frac 1 a \leq 1$ and $\frac 1 b \leq 1$ so $\frac 1 a +\frac 1 b\leq 2$. If $\frac 1 a +\frac 1 b$ is an integer then it must be $1$ or $2$. Also, if $a>2$ or $b >2$ we get $\frac 1 a +\frac 1 b <1.5 <2$ so we mjust have $a \leq 2$ and $b \leq 2$. You can now finish the proof easily.

Answer 2

I want to suggest much easier approach:

You only need to check few cases:

Case 1: If $a=1$ then $\frac{1}{b}$ must be an integer which is only possible if $b=1$.

Case 2: If $a=2$ then $\frac{1}{b} = k+\frac{1}{2}$ but clearly $0\leq \frac{1}{b}\leq 1$ so $\frac{1}{b}=\frac{1}{2}$ hence $b=2$.

Case 3: If $a\geq 3$ then $\frac{1}{a}\leq \frac{1}{3}$ and so $\frac{1}{b}\geq \frac{2}{3}$ but there is no natural number $b$ satisfying this condition other than $b=1$ in this case $a=1$ as in Case 1. (because $\frac{1}{x}$ is decreasing and $\frac{1}{2}$ is already smaller).

Answer 3

Write $a=dx$ and $b=dy$ where $d=\gcd(a,b)$ and $x,y$ are relatively prime. Then

$$ab\mid a+b\implies d^2xy\mid d(x+y) \implies dxy\mid x+y \implies x\mid x+y$$ $$\implies x\mid (x+y)-x \implies x\mid y \implies x=1$$

With the same logic we get $y=1$.

Answer 4

On the one hand: $$b+a=kab \Rightarrow (ak-1)(bk-1)=1 \Rightarrow \\ \begin{cases}ak-1=1 \\ bk-1=1\end{cases} \Rightarrow \begin{cases}a=\frac 2k\ge 1 \\ b=\frac 2k\ge 1\end{cases} \Rightarrow k\le 2.$$ On the other hand: $$a,b\ge 1 \Rightarrow \frac1a +\frac1b=k>0.$$

Thus: $k=1$ and $k=2$.

Answer 5

WLOG $a\le b$.

$$\begin{array}{c|ccc} &a=1&a=2&a>2 \\\hline b=1&\color{green}1&-&- \\b=2&\frac32&\color{green}1&- \\b>2&1<1+\frac1b<2&0<\frac12+\frac1b<1&0<\frac1a+\frac1b<1 \end{array}$$