Asian Pacific Mathematics Olympiad

By inspection, we see that the pairs $(a, b)$ with $a = b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.

- Case 1. Consider the case $b < a$. Since $b-1$ is a multiple of $a-1$, it follows that $b = 1$. This yields the second set of solutions described above.

- Case 2. This leaves the case $b \geq a$. Since the positive integer $a^{3}$ is a multiple of $b^{2}$, there is a positive integer $c$ such that $a^{3} = b^{2} c$.

Note that $a \equiv b \equiv 1$ modulo $a-1$. So we have
$$1 \equiv a^{3} = b^{2} c \equiv c \quad (\bmod a-1)$$
If $c < a$, then we must have $c = 1$, hence, $a^{3} = b^{2}$. So there is a positive integer $d$ such that $a = d^{2}$ and $b = d^{3}$. Now $a-1 \mid b-1$ yields $d^{2}-1 \mid d^{3}-1$. This implies that $d+1 \mid d(d+1)+1$, which is impossible.

If $c \geq a$, then $b^{2} c \geq b^{2} a \geq a^{3} = b^{2} c$. So there's equality throughout, implying $a = c = b$. This yields the first set of solutions described above.

Therefore, the solutions described above are the only solutions.

We will start by showing that there are positive integers $x, c, d$ such that $a = x^{2} c d$ and $b = x^{3} c$. Let $g = \operatorname{gcd}(a, b)$ so that $a = g d$ and $b = g x$ for some coprime $d$ and $x$. Then, $b^{2} \mid a^{3}$ is equivalent to $g^{2} x^{2} \mid g^{3} d^{3}$, which is equivalent to $x^{2} \mid g d^{3}$. Since $x$ and $d$ are coprime, this implies $x^{2} \mid g$. Hence, $g = x^{2} c$ for some $c$, giving $a = x^{2} c d$ and $b = x^{3} c$ as required.

Now, it remains to find all positive integers $x, c, d$ satisfying
$$x^{2} c d - 1 \mid x^{3} c - 1$$
That is, $x^{3} c \equiv 1 \left(\bmod x^{2} c d - 1\right)$. Assuming that this congruence holds, it follows that $d \equiv x^{3} c d \equiv x \left(\bmod x^{2} c d - 1\right)$. Then, either $x = d$ or $x - d \geq x^{2} c d - 1$ or $d - x \geq x^{2} c d - 1$.

- If $x = d$ then $b = a$.
- If $x - d \geq x^{2} c d - 1$, then $x - d \geq x^{2} c d - 1 \geq x - 1 \geq x - d$. Hence, each of these inequalities must in fact be an equality. This implies that $x = c = d = 1$, which implies that $a = b = 1$.
- If $d - x \geq x^{2} c d - 1$, then $d - x \geq x^{2} c d - 1 \geq d - 1 \geq d - x$. Hence, each of these inequalities must in fact be an equality. This implies that $x = c = 1$, which implies that $b = 1$.

Hence the only solutions are the pairs $(a, b)$ such that $a = b$ or $b = 1$. These pairs can be checked to satisfy the given conditions. $\square$

All answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.

To show that these are all solutions, note that we can easily eliminate the case $a = 1$ or $b = 1$. Thus, assume that $a, b \neq 1$ and $a \neq b$. By the second divisibility, we see that $a-1 \mid b-a$. However, $\operatorname{gcd}(a, b) \mid b-a$ and $a-1$ is relatively prime to $\operatorname{gcd}(a, b)$. This implies that $(a-1) \operatorname{gcd}(a, b) \mid b-a$, which implies $\operatorname{gcd}(a, b) \left\lvert\, \frac{b-1}{a-1} - 1\right.$.

The last relation implies that $\operatorname{gcd}(a, b) < \frac{b-1}{a-1}$, since the right-hand side are positive. However, due to the first divisibility,
$$\operatorname{gcd}(a, b)^{3} = \operatorname{gcd}\left(a^{3}, b^{3}\right) \geq \operatorname{gcd}\left(b^{2}, b^{3}\right) = b^{2} .$$
Combining these two inequalities, we get that
$$b^{\frac{2}{3}} < \frac{b-1}{a-1} < 2 \frac{b}{a} .$$
This implies $a < 2 b^{\frac{1}{3}}$. However, $b^{2} \mid a^{3}$ gives $b \leq a^{\frac{3}{2}}$. This forces
$$a < 2\left(a^{\frac{3}{2}}\right)^{\frac{1}{3}} = 2 \sqrt{a} \Longrightarrow a < 4 .$$
Extracting $a = 2, 3$ by hand yields no additional solution. $\square$