Problem 1433

To show that the set $\{p_i\}_{i \in \mathbb{N}}$ is finite, we will use induction and properties of prime numbers.

1. Base Case:
Consider a positive integer $k$ such that $\max(p_1, p_2) \le k \cdot 2021! + 1$. This is our initial assumption.

2. Inductive Step:
Assume that for some $n$, $p_n \le k \cdot 2021! + 1$ and $p_{n+1} \le k \cdot 2021! + 1$. We need to show that $p_{n+2} \le k \cdot 2021! + 1$.

- **Case 1: $p_{n} = 2$ or $p_{n+1} = 2$:**
If either $p_n$ or $p_{n+1}$ is 2, then:
$$p_{n+2} \le (k \cdot 2021! + 1) + 2 + 2018 = k \cdot 2021! + 2021$$
However, since $k \cdot 2021! + 1, \ldots, k \cdot 2021! + 2020$ are all composite (as they are consecutive integers greater than $2021!$), the largest prime divisor $p_{n+2}$ must be less than or equal to $k \cdot 2021! + 1$.

- **Case 2: $p_{n} > 2$ and $p_{n+1} > 2$:**
Since $p_n$ and $p_{n+1}$ are both odd primes, $p_n + p_{n+1} + 2018$ is even. Therefore, the largest prime divisor $p_{n+2}$ must be less than or equal to half of this sum:
$$p_{n+2} \le \frac{p_n + p_{n+1} + 2018}{2} \le \frac{(k \cdot 2021! + 1) + (k \cdot 2021! + 1) + 2018}{2} = k \cdot 2021! + 1009$$
Again, since $k \cdot 2021! + 1, \ldots, k \cdot 2021! + 2020$ are all composite, the largest prime divisor $p_{n+2}$ must be less than or equal to $k \cdot 2021! + 1$.

3. Conclusion:
By induction, we have shown that $p_i \le k \cdot 2021! + 1$ for all $i$. Since there are only finitely many primes less than or equal to $k \cdot 2021! + 1$, the set $\{p_i\}_{i \in \mathbb{N}}$ must be finite.

$\blacksquare$