Htykuut

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 $dfrac{b}{a} + dfrac{c}{b} + dfrac{a}{c} in mathbb{Z}$.

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 $pm1$.