1. Problem: Find all integers $a, b$ such that
$$
a^{2}+b=b^{2022} \text {. }
$$
Solution: (Ethan $\mathrm{Ng}$ )
Let $g=\operatorname{gcd}\left(b, b^{2021}-1\right)$. Since we cannot have both $b$ and $\left(b^{2021}-1\right)$ being zero, $g$ must be a positive integer. Since $g \mid b$ we must have $g \mid b^{2021}$ and therefore
$$
g \mid\left(b^{20221}\right)-\left(b^{2021}-1\right)=1 .
$$

Hence $g=1$ and thus $b$ and $\left(b^{2021}-1\right)$ are coprime. Since the product $b \times\left(b^{2021}-1\right)=$ $b^{2022}-b=a^{2}$ is a perfect square, we get both factors $b$ and $\left(b^{2021}-1\right)$ must be perfect squares or the negatives of perfect squares, or one of them must be zero.
- If both $b$ and $\left(b^{2021}-1\right)$ are positive perfect squares, then let $b=x^{2}$ and $\left(b^{2021}-1\right)=$ $y^{2}$. Therefore $\left(x^{2}\right)^{2021}-1=y^{2}$ and thus
$$
1=x^{4042}-y^{2}=\left(x^{2021}-y\right)\left(x^{2021}+y\right) \text {. }
$$

However since 1 is prime and the only integer factorizations of 1 are $1 \times 1$ and $(-1) \times(-1)$, we must have
$$
\left(x^{2021}-y\right)=\left(x^{2021}+y\right)
$$

Hence $y=0$ which is a contradiction.
- If both $b$ and $\left(b^{2021}-1\right)$ are negative perfect squares, then let $b=-x^{2}$ and $\left(b^{2021}-1\right)=$ $-y^{2}$. Therefore $\left(-x^{2}\right)^{2021}-1=-y^{2}$ and thus
$$
x^{4042}+y^{2}=1 \text {. }
$$

However the minimum value of $x^{4042}+y^{2}$ is $1+1=2$, so this is a contradiction too.
- If $b=0$ then we get $a^{2}+0=0^{2022}$. So $a=0$ and we get the solution $(a, b)=(0,0)$.
- If $\left(b^{2021}-1\right)=0$ then we get $b=1$. So $a^{2}+1=1^{2022}$. So $a=0$ and we get the solution $(a, b)=(0,1)$.
Therefore the only solutions for $(a, b)$ are: $(0,0)$ and $(0,1)$.