Reciprocals of integers

The only possible solutions are ones where $a=b$ . Notice that every possible solution (a,b,c) can be written as (kd,ke,c) where d and e are coprime. Therefore, we will assume that d and e are coprime, then at the end scale d and e up by all positive integers k. $\frac ed+\frac de= \frac{d^2+e^2}{de}=\frac{(d+e)^2}{de}-2$ For this to be an integer, $de$ must divide $(d+e)^2$ . Because d and e are squarefree and coprime, de is squarefree. This means that $\frac{(d+e)^2}{de}$ will be an integer only when $\frac{d+e}{de}=\frac 1d+\frac 1e$ is an integer. This is only true when $d=e=1$ . Scaling this solution by all squarefree positive integers tells us that $a=b$ where a and b are any squarefree positive integer is the only possible solution to the equation and $a-b=0$ .

Note: This is still true even if a and b are not squarefree. We can show this by Vieta root jumping. If (a,b) is a solution, WLOG say that $a \geq b$ Either a=b, or $a>b$ and a is a root of $x^2-cbx+b^2$ . By Vieta, this equation will have another root at $r=cb-a=\frac {b^2}a$ . Because this is an integer and greater than 0, $(b,\frac {b^2}a)$ will be another solution where b is greater than $\frac {b^2}a$ . Repeating this process will give us a list of decreasing numbers that all satisfy $\frac{r_n}{r_{n+1}}+\frac{r_{n+1}}{r_n}=c$ . Eventually one of the numbers will reach 1 which shows that c=2 and a=b. This contradicts our original assumption that a>b, therefore we know that a must equal b.

(1). Suppose that there exist distinct positive integers $a$ and $b$ such $\frac{b}{a} + \frac{a}{b}$ is an integer $c$ . Let $\frac{a}{b} = p$ . Then $p + \frac{1}{p} = c\implies p^2 -pc +1 =0$ Solving for $p$ $\ p = \dfrac{c \pm\sqrt{c^2-4}}{2}$ $p$ is an integer if $\sqrt{c^2-4}$ is perfect square number for $c> 1$ and $c$ must be an even integer say $c=2k$ that is $p = \dfrac{2k\pm\sqrt{4k^2-4}}{2} = k\pm\sqrt{k^2-1}$ $\sqrt{k^2-1}$ is only a perfect square number iff $\sqrt{k^2-1} =0$ impliging that $c=2$ for $k=1$ finally results $p + \frac{1}{p} = 2$ or $p = 1 \implies \frac{a}{b} = 1$ . Therefore, $a =b$ which is in contradiction to our supposition. Now $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{2}{a} = c$ shows that $c$ is an integer either $a = 1$ or $a=2$ . Hence, $a-b =0$

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(2). $\dfrac{b}{a} + \dfrac{a}{b} = \dfrac{1}{a} + \dfrac{1}{b} = c$ First we note that $a$ and $b$ are positive as well as square free numbers. Being square free $a$ and $b$ can be either co-prime numbers or $\text{g.c d}(a,b)$ can be $a$ or $b$ .

If $a$ and $b$ are co-prime then We cleary can notice that $ab > a+b$ for any positive integers $(a> 1$ or $b>1$ ) $\dfrac{a +b}{ab} = \dfrac{p}{q} \neq c \phantom{ccc} {\color{#3D99F6}\dfrac{p}{q} < 1}$

If $\text{g.c.d}(a,b)$ is either $a$ or $b$ and $a > b / a< b$ then $a = bk$ or $b =ak$ where $k$ and $b$ or $k$ and $a$ are only prime factors of $a$ . $\dfrac{1}{a}+ \dfrac{1}{b} = \dfrac{1}{bk}+ \dfrac{1}{b} = \dfrac{k+1}{bk} = \dfrac{p_0}{q_0} \neq c \phantom{ccc} {\color{#3D99F6}\dfrac{p_0}{q_0} < 1}$ Since $bk> k+1\implies p_0 < q_0$ . It is clear that $c$ can be only be integers iff $a=b$ & $0 < (a,b)\leq 2$

Therefore, $c$ can reduced to integer solution only for $a =b = 1$ or $a =b =2$ . So the possible solution for $a-b =\boxed{0}$

Since a and b are squarefree,the maximum value of c is $\frac{1}{2}+\frac{1}{2}=1$ (when a and b are both 2).If we change the value of a and b,then c will be smaller(and of course c will not be a positive integer).So,a-b=2-2=0