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

  • 3
    $\begingroup$ Surely, one solution is obvious. $\endgroup$ – user608030 Dec 10 '18 at 13:10
  • $\begingroup$ 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. $\endgroup$ – hellHound Dec 10 '18 at 13:22
  • $\begingroup$ @hellhound it does not satisfy $b/a + c/b + a/c \in \mathbb{Z}$, only $a/b+b/c+c/a \in \mathbb{Z}$ $\endgroup$ – ShakesBeer Dec 10 '18 at 13:25
  • $\begingroup$ 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. $\endgroup$ – hellHound Dec 10 '18 at 13:27
  • 1
    $\begingroup$ @hellhound correct, none of them do :) $\endgroup$ – ShakesBeer 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 $\dfrac ab, \dfrac bc, \dfrac ca \in \{-1,1\}$.

  • 1
    $\begingroup$ Beautiful! I don't think I ever would have thought of this. $\endgroup$ – ShakesBeer Dec 10 '18 at 15:27
  • $\begingroup$ @Shakespeare See also this prior symmetric variant. $\endgroup$ – Bill Dubuque Dec 10 '18 at 22:21

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 $\pm1$.


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