Find $a, b \in \mathbb{N} $ such that $\frac{1}{a}+\frac{1}{b} +\frac{1}{(a,b)}+\frac{1}{[a,b]}=\frac{1}{2}$

Question

Find $a, b \in \mathbb{N} $ such that $$\frac{1}{a}+\frac{1}{b} +\frac{1}{(a,b)}+\frac{1}{[a,b]}=\frac{1}{2}.$$

My try :

$$\frac{[a,b]+(a,b)}{[a,b] \cdot (a,b)}=\frac{ab-2a-2b}{2ab}.$$

Now what to do?

Answer 1

Let $gcd(a,b)=d$. Thus, $$\frac{1}{a}+\frac{1}{b}\geq\frac{1}{(a,b)}+\frac{1}{[a,b]}$$ it's $$\frac{1}{a_1d}+\frac{1}{b_1d}\geq\frac{1}{d}+\frac{1}{da_1b_1}$$ or $$(a_1-1)(b_1-1)\geq0,$$ which is obvious.

Now, let $a\geq b$.

Thus, $$\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{(a,b)}+\frac{1}{[a,b]}\leq2\left(\frac{1}{a}+\frac{1}{b}\right)\leq\frac{4}{b},$$ which gives $b\leq8$, which says $d\leq8$.

Also, $$\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{(a,b)}+\frac{1}{[a,b]}$$ it's $$\frac{1}{a_1d}+\frac{1}{b_1d}+\frac{1}{d}+\frac{1}{da_1b_1}=\frac{1}{2}$$ or $$2(1+a_1)(1+b_1)=a_1b_1d,$$ which gives $d>2$.

Now, it remains a work with last equation for $3\leq d\leq8,$ which gives all solutions.

Can you end it now?

Answer 2

Given $(c,d)=1$ and $g$, we get the pair $(a,b)=(gc,gd)$ and $$ \begin{align} \frac12 &=\frac1a+\frac1b+\frac1{(a,b)}+\frac1{[a,b]}\\ &=\frac1{gc}+\frac1{gd}+\frac1g+\frac1{gcd}\\ &=\frac1g\left(1+\frac1c\right)\left(1+\frac1d\right) \end{align} $$ That is $$ g=2\left(1+\frac1c\right)\left(1+\frac1d\right) $$ Here are all the possibilities $$ \begin{array}{c|c|c|c} c&d&g&(gc,gd)\\\hline 1&1&8&(8,8)\\ 1&2&6&(6,12)\\ 1&4&5&(5,20)\\ 2&3&4&(8,12)\\ 3&8&3&(9,24)\\ 4&5&3&(12,15) \end{array} $$