Japan competition problems, 2022

For a prime $p$ and a positive integer $n$, denote by $\text{ord}_p n$ the largest nonnegative integer $i$ such that $p^i$ divides $n$.

First we consider the case $y \equiv 0 \pmod 2$. Since $3^z = 8^y + 2xy + 1 \equiv 1 \pmod 4$, $x$ is even. Write $x = 2z$ and $y = 2w$ for positive integers $z$ and $w$, and we have $8zw + 1 = 3^{2z} - 8^{2w} = (3^z - 8^w)(3^z + 8^w) \ge 3^z + 8^w$. $3^z - 8^w > 0$ shows $z > w$, hence we have $8z^2 + 1 > 8zw + 1 > 3^z$. If $z \ge 6$, by the binomial theorem we have
$$3^z = (2+1)^z \ge 2^{z-2} \cdot {}_zC_2 + 2^{z-1}z + 2^z > 16 \cdot {}_zC_2 + 32z + 1 > 8z^2 + 1,$$
which is absurd. If $z = 5$, $3^z > 8z^2 + 1$ holds. It follows that $z = 1, 2, 3, 4$ is necessary. Since $9^4 \ge 9^z > 64^w$, $w = 1, 2$ is necessary. Moreover, $3^z + 8^w \le 8zw + 1 \le 65$ shows that $w = 1$ is necessary. Among $z = 1, 2, 3, 4$, only $z = 2$ satisfies $9^z = 8z + 65$. Therefore the solution in this case is $(z, w) = (2, 1)$, that is, $(x, y) = (4, 2)$.

Next we consider the case $y \equiv 1 \pmod 2$. From $8^y + 1 = 3^z - 2xy$ we have $\text{ord}_3(8^y + 1) = \text{ord}_3(3^z - 2xy)$. Here we use the following lemma:
Lemma. For any positive odd integer $y$, $\text{ord}_3(8^y + 1) = \text{ord}_3 y + 2$.
Proof. We proceed by induction on $y$. If $y = 1$, $\text{ord}_3(8^y + 1) = \text{ord}_3 y + 2 = 2$, thus the claim is true. Let $\ell \ge 3$ be an odd integer and assume that the claim is true for $y < \ell$.
If $\ell$ is not a multiple of 3, we can write $\ell = 3k+r$ with a nonnegative integer $k$ and $r \in \{1, 2\}$. Then we have $8^\ell = 8^r \times 512^k \equiv (-1)^k 8^r \pmod{27}$. Since $r \in \{1, 2\}$, $8^\ell + 1 \not\equiv 0 \pmod{27}$. On the other hand, since $\ell$ is odd we have $8^\ell + 1 \equiv 0 \pmod 9$, thus $\text{ord}_3(8^\ell + 1) = 2$. Therefore the claim is true when $\ell$ is not a multiple of 3.
If $\ell$ is a multiple of 3, we can write $\ell = 3k$ for a positive odd integer $k$. Then we have $\text{ord}_3(8^\ell + 1) = \text{ord}_3(8^k + 1) + \text{ord}_3(64^k - 8^k + 1)$, for $8^\ell + 1 = (8^k + 1)(64^k - 8^k + 1)$. Since $k < \ell$, inductive hypothesis shows that $\text{ord}_3(8^k + 1) = \text{ord}_3 k + 2$. From $64^k - 8^k + 1 \equiv 3 \pmod 9$ we have $\text{ord}_3(64^k - 8^k + 1) = 1$. It follows that $\text{ord}_3(8^\ell + 1) = \text{ord}_3 k + 3 = \text{ord}_3 \ell + 2$. Therefore the claim is also true when $\ell$ is a multiple of 3 and the induction is complete. $\blacksquare$

Since $0 < 3^z - 2xy < 3^z$ we have $\text{ord}_3(3^z - 2xy) < x$. This and the lemma show that $\text{ord}_3 2xy = \text{ord}_3(3^z - 2xy) = \text{ord}_3 y + 2$, thus we have $\text{ord}_3 x = 2$. Hence $x$ is a multiple of 9 and can be written as $x = 3v$ where $v$ is a multiple of 3. Then we have $6vy + 1 = 3^{3v} - 2^{3v} = (3^v - 2^v)(9^v + 3^v \cdot 2^v + 4^v) > 9^v$. From $3^v - 2^v > 0$, we have $2v > y$ and $12v^2 + 1 > 6vy + 1 > 9^v$ holds. Since $v \ge 3$, the binomial theorem shows that
$$9^v = (8+1)^v \ge 64 \cdot {}_vC_2 + 8v + 1 > 12v^2 + 1,$$
which is a contradiction. As a result, there exist no solutions when $y$ is odd.
From above, the answer is $(x, y) = (4, 2)$.