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

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.

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.

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.

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!

• Suggested edit to title: dividing each by (Slightly clearer, and its not worth me submitting a suggested 1-word edit for review!) – timtfj Dec 7 '18 at 22:49
• I can't edit this myself; 15/24 is not irreducible. – RandomMathGuy Dec 7 '18 at 22:56
• @RandomMathGuy Thanks for pointing that out! (Also clarified the title, thank you timtfj.) – orion2112 Dec 7 '18 at 22:59

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.
• The smallest counterexample is $2,3,25$. An infinite family of counterexamples is $2,3,25+30n$. – lhf Dec 7 '18 at 23:58
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*}