1. Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.
Solution
Solution 1.1
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.

Solution 1.2
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
$$
x^{2} c d-1 \mid x^{3} c-1 \text {. }
$$

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.
Solution 1.3
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.