Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction

Question

Let $a$, $b$ and $c$ be $3$ integers with no common factors.

I conjecture that at least one of the three fractions:

$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$

is irreducible.

I know that they are not necessarily all irreducible.

For instance, taking $(a,b,c)=(2,7,15)$, it is verified that $2$, $7$ and $15$ have no common factors.

We have $a+b+c=24$ so $$\dfrac{2}{2+7+15}=\dfrac{2}{24}\quad(\text{reducible)}$$ $$\dfrac{7}{2+7+15}=\dfrac{7}{24}\quad(\text{irreducible)}$$ $$\dfrac{15}{2+7+15}=\dfrac{15}{24}\quad(\text{reducible)}$$

I have yet to find any counter-example to the initial statement so I'm pretty certain it must be true. I believe I managed to prove the analogous statement for $2$ integers but I would like to generalize it to $3$, and eventually $n$, integers.

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Proof for two integers $a$ and $b$ with no common factors

Suppose $k$ is a factor of $a$ and that $a'$ is an integer such that $a=ka'$. Then:

$$\frac{a}{a+b}=\frac{ka'}{ka'+b}=\frac{ka'}{k\left(a'+\dfrac{b}{k}\right)}=\dfrac{a'}{a'+\dfrac{b}{k}}$$

But $k$ is a factor of $a$ and $a$ and $b$ have no common factors by hypothesis, so $k$ does not divide $b$. So $\dfrac{b}{k}$ is irreducible. Since $a'$ is an integer and $\dfrac{b}{k}$ isn't (assuming $k\neq 1$), then $\dfrac{a'}{a'+\dfrac{b}{k}}$ is not a 'fraction' in the sense that it is not a ratio of two integers, making the original fraction $\dfrac{a}{a+b}$ irreducible.

This proves (correct me if I'm wrong) that at least one of the the fractions $\dfrac{a}{a+b}$ or $\dfrac{b}{a+b}$ is irreducible.

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For the case with $3$ integers, a similar argument doesn't seem to work. Taking $(4,5,11)$ as an example:

$$\frac{4}{4+5+11}=\frac{4}{4\left(1+\dfrac{5}{4}+\dfrac{11}{4}\right)}=\frac{1}{1+\dfrac{16}{4}}=\frac{1}{5}$$

the fact that there are now $3$ numbers allows for some fractions to combine into an integer. Because of this, I'm having difficulty proving the statement further.

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I'm not particularly good with number theory proofs and I apologize if this is not as elegant as it should be (or worse, if there are mistakes). I'm welcoming any tips and possible improvements. Thanks in advance!

Answer 1

Counterexample: $5,7,58$. \begin{align*}\frac{5}{5+7+58} &= \frac{5}{70} = \frac1{14}\\ \frac{7}{5+7+58} &= \frac{7}{70} = \frac1{10}\\ \frac{58}{5+7+58} &= \frac{58}{70} = \frac{29}{35}\end{align*}

How did I find it? Choose $a$ and $b$ odd and relatively prime, then let $c=2ab-a-b$. This $c$ must be even, which makes $\frac{c}{a+b+c}$ reducible.

Answer 2

The smallest counterexample is $(2,3,25)$. An infinite family of counterexamples is $(2,3,25+30n)$: $$ \begin{align*} \frac{a}{a+b+c} &= \frac{2}{30+30n} = \frac{1}{15+15n} \\ \frac{b}{a+b+c} &= \frac{3}{30+30n} = \frac{1}{10+10n} \\ \frac{c}{a+b+c} &= \frac{25+30n}{30+30n} = \frac{5+6n}{6+6n} \end{align*} $$

There are several other infinite family of counterexamples.