Find all the integers which are of form $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}, a,b,c\in \mathbb{N}$, any two of $a,b,c$ are relatively prime.

Question

I have a question which askes to find all the integers which can be expressed as

$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}$

where $a,b,c\in \mathbb{N} $ and any two of $a,b,c$ are relatively prime.

My approach:

Since they told any two of $a,b,c$ are relatively prime, so:

$\displaystyle \tag*{} \begin{align} \text{gcd}(a,b) &= 1 \\ \text{gcd}(b,c) &= 1 \\ \text{gcd}(c,a) &= 1\end{align}$

We have: $\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c} = \dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}$

And we have to prove: $\displaystyle \tag*{} \frac{a}{b} + \frac{b}{a}+ \frac{c}{a} + \frac{a}{c}+ \frac{b}{c} + \frac{c}{b} \in \mathbb{N} $

We have the following that if $\text{gcd}(m,n)=1$ and $m,n,k \in \mathbb{Z}$ then

$\displaystyle \tag*{} \frac{m}{n} + \frac{n}{m} = k$

has only $1$ positive solution, which is $m=n=1$

So, the only solution I found is $a=b=c=1$.

Is there anything I am missing? Any help would be appreciated, thank you. :)

Answer 1

$\frac{ab(a+b)+bc(b+c)+ca(c+a)}{abc}\in \mathbb{N}$, therefore $abc|ab(a+b)+bc(b+c)+ca(c+a)$. Because $ab(a+b)$ and $ca(c+a)$ is a multiple of $a$, $a|bc(b+c)$ is true. $bc$ is coprime with $a$, thus $a|b+c$. Using the same ways, it can be proven that $$a|b+c \\ b|a+c \\ c|a+b$$ WLOG $a\geq b\geq c$. Therefore $b+c$ is $a$ or $2a$. If $a=b+c$, then $b|b+2c$ and $b|2c$. If $b=c$, then $(a,b,c)$ is $(2,1,1)$ and the value is 7. If $b=2c$, $(a,b,c)$ is $(3,2,1)$ and the value is 8. If $2a=b+c$, then $(a,b,c)$ is $(1,1,1)$ and the value is 6. Therefore, all integers possible are 6, 7 and 8.

Answer 2

NB: While I was typing up my proof, @user provided a similar proof, leaving mine essentially a duplicate. However, I'll leave it since I may have written it out in slightly greater detail.

First, we may note that $c$ must divide $a+b$, etc. To see this, write the sum on common denominator $$ \frac{b+c}a+\frac{a+c}b+\frac{a+b}c =\frac{bc(b+c)+ac(a+c)+ab(a+b)}{abc} $$ and note that, for $c$ to divide the numerator, after removing terms that are multiples of $c$, we are left with $c|ab(a+b)$, which forces $c|a+b$. Similarly, $a|b+c$ and $b|a+c$.

We may assume that $a\le b\le c$, where equality is only possible for the number 1. So let's first deal with the cases where equalities are an issue.

First, $a=b=c=1$ is a solution. Ie, $(a,b,c)=(1,1,1)$.

Next, if $a=b=1<c$, we need $c|a+b$, for which the only option is $c=2$. This is also a solution. Ie $(a,b,c)=(1,1,2)$.

So now we are left with the case $a<b<c$. Since $c|a+b$, but $a+b<2c$, we must have $c=a+b$. Next $b|a+c=2a+b$, so $b|2a$; however, $a<b$, so $2a=b$ since $2a$ is a multiple of $b$ but less than $2b$. This also forces $a=1$ since $a$ and $b=2a$ should be relatively prime. This makes $(a,b,c)=(1,2,3)$.