Bay Area Mathematical Olympiad

Solution:

a.
For example, $(1, 2, 4, 5)$ and $(9, 2, 12, 6)$ work.

b.
Before solving the problem, we establish a useful definition and lemma.

If $p$ is a prime and $x$ is a nonzero rational number, we define $\operatorname{ord}_p x$ to be the unique integer $k$ such that $p^{-k} x$ is an integer not divisible by $p$. For example, $\operatorname{ord}_3 45 = 2$ and $\operatorname{ord}_5 \frac{11}{5} = -1$. Then we claim the following:

Lemma. Let $p$ be prime and let $x, y$ be nonzero rational numbers. Then:

(i) $\operatorname{ord}_p(xy) = \operatorname{ord}_p x + \operatorname{ord}_p y$.

(ii) $\operatorname{ord}_p(x / y) = \operatorname{ord}_p x - \operatorname{ord}_p y$.

(iii) If $\operatorname{ord}_p x < \operatorname{ord}_p y$, then $\operatorname{ord}_p(x + y) = \operatorname{ord}_p x$.

Proof of the lemma. Parts (i)-(ii) are straightforward. For part (iii), let $k = \operatorname{ord}_p x$ and $k + \ell = \operatorname{ord}_p y$, where we assume $\ell > 0$. Then $p^{-k} x = n$ and $p^{-k-\ell} y = m$ for some integers $m, n$ not divisible by $p$. It follows that $p^{-k}(x + y) = n + p^{\ell} m$, and this is an integer not divisible by $p$. Thus $\operatorname{ord}_p(x + y) = k$ as claimed, proving the lemma.

Now we turn to the problem.

Suppose $a, b, c$ are integers such that $a / b + b / c + c / a$ is an integer. We will show that $\operatorname{ord}_p(a b c)$ is a multiple of $3$ for all primes $p$, which implies that $a b c$ is a perfect cube.

Let $p$ be a prime and let $r = \operatorname{ord}_p a$, $s = \operatorname{ord}_p b$, $t = \operatorname{ord}_p c$. By part (i) of the lemma, $\operatorname{ord}_p(a b c) = r + s + t$.

By part (ii) of the lemma, we have $\operatorname{ord}_p(a / b) = r - s$, $\operatorname{ord}_p(b / c) = s - t$, and $\operatorname{ord}_p(c / a) = t - r$. If these three differences are $0$, then $r = s = t$ and $r + s + t$ is trivially a multiple of $3$. Otherwise, the least of $r - s, s - t, t - r$ is negative and the greatest is positive (since their sum is $0$).

We now claim that at least two of $r - s, s - t, t - r$ must be tied for least. If this is not true, then, by applying part (iii) of the lemma, we may conclude that $\operatorname{ord}_p(a / b + b / c + c / a) = \min \{ r - s, s - t, t - r \} < 0$, which contradicts the assumption that $a / b + b / c + c / a$ is an integer. Thus we have proven our claim. But if two of $r - s, s - t, t - r$ are equal, then $r, s, t$ (in some order) form an arithmetic progression, and $r + s + t$ is three times the middle term of the progression. Thus we have shown that $r + s + t$ is a multiple of $3$; in other words, $\operatorname{ord}_p(a b c)$ is a multiple of $3$ for all primes $p$, and we are finished.