# Solutions to $a,\ b,\ c,\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}$

I came across a puzzle in a Maths Calendar I own. Most of them I can do fairly easily, but this one has me stumped, and I was hoping for a hint or solution. The question is:

What are the solutions to

$$\left \{ a,\ b,\ c,\ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a},\ \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \right \} \subset \mathbb{Z}$$

I've tried a few things, but don't think I've made any meaningful progress, besides determining that $$a = \pm b = \pm c \$$ are the only obvious possible solutions. My hope is to prove that no other solution can exist.

I don't know if it helps, but I also did a brute force search for coprime numbers $$a,b,c$$ for which $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \in \mathbb{Z}$$, with $$1 \leq a \leq b \leq c$$, and $$a \leq 100, b \leq 1000, c \leq 10000$$.

The reason for coprimality is that if a solution has a common factor, we can divide through by the common factor and have another solution that satisfies the conditions.

The triplets I found which satisfy this are:

$$(a, b, c) = (1, 1, 1), (1,2,4), (2, 36, 81), (3, 126, 196), (4, 9, 162), (9, 14, 588), (12, 63, 98), (18, 28, 147), (98, 108, 5103)$$

None of these except the first satisfy $$\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \in \mathbb{Z}$$.

• Surely, one solution is obvious. – user608030 Dec 10 '18 at 13:10
• How is $(a,b,c)=(1,2,4)$ a solution? $\frac{b}{a} + \frac{c}{b} + \frac{a}{c} = \frac{2}{1} + \frac{4}{2} + \frac{1}{4} = 4 + 1/4$, not an integer. – hellHound Dec 10 '18 at 13:22
• @hellhound it does not satisfy $b/a + c/b + a/c \in \mathbb{Z}$, only $a/b+b/c+c/a \in \mathbb{Z}$ – Shakespeare Dec 10 '18 at 13:25
• Oh, I noticed the line about your search now, my bad. Although, if I had to guess, none of the triples from your search except $(1,1,1)$ would satisfy the problem's requirement. – hellHound Dec 10 '18 at 13:27
• @hellhound correct, none of them do :) – Shakespeare Dec 10 '18 at 13:28

Suppose that $$\displaystyle a,b,c,\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \in \mathbb Z$$. Consider polynomial $$P(x)=\left(x-\frac{a}{b}\right)\left(x-\frac{b}{c}\right)\left(x-\frac{c}{a}\right) = x^3-\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)x^2+\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)x-1.$$ Its coefficients are integers. Since the leading coefficient is $$1$$, all rational roots of $$P$$ are integers. Since the constant term is $$-1$$, it follows that all integer roots of $$P$$ are $$1$$ or $$-1$$ (they must divide the constant term). Since $$\dfrac ab, \dfrac bc, \dfrac ca$$ are rational roots of $$P$$, it follows that $$\dfrac ab, \dfrac bc, \dfrac ca \in \{-1,1\}$$.
Let $$(a,b,c)$$ satisfy the requirements. Let $$(a,b,c)$$ are coprime. Then $$abc$$ divides both $$a^2c + b^2a + c^2b$$ and $$a^2b+b^2c+c^2a$$.
Let $$p$$ be a prime factor of $$a$$. Let $$d$$ be the largest number such that $$p^d$$ divides $$a$$. Then $$p$$ divides $$b^2c$$ and $$c^2b$$. Assume $$p$$ divides $$b$$ (and does not divide $$c$$).
Since $$p^{d+1}$$ divides $$a^2$$, $$ab$$, and $$abc$$, where the latter divides $$a^2c + b^2a + c^2b$$, it follows $$p^{d+1}$$ divides $$b$$. This in turn implies that $$p^{d+1}$$ divides $$a^2b+b^2c+c^2a$$. Now, $$p$$ does not divide $$c$$ by assumption of coprimality, hence $$p^{d+1}$$ divides $$a$$, a contradiction to the maximality of $$d$$.
Hence, none of $$a,b,c$$ has a prime factor. So all these numbers are equal $$\pm1$$.