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Problem

Let \( (p_i) \) be a sequence of prime numbers such that \(p_n\) is the largest prime divisor of $$ p_{n-1}+p_{n-2}+2000, \quad \forall n. $$ Prove that the sequence \( (p_n) \) is bounded.

Solution

Let

$$ P_n = \max\{p_1,p_2,\ldots,p_n\}. $$

Since \(p_{n-1}\) and \(p_{n-2}\) are odd primes, we have

$$ p_n \le \frac{p_{n-1}+p_{n-2}+2000}{2} \le \max\{p_{n-1},p_{n-2}\}+1000 = P_{n-1}+1000. $$

Moreover, if the prime \(2\) appears, then

$$ p_n \le P_{n-1}+2002. \tag{1} $$

From the two estimates above, we deduce that

$$ p_n \le P_{n-1}+2002. $$

Lemma. For every \(n\in\mathbb{N}\), there exist \(n\) consecutive composite numbers.

Choose \(k>2026\). Then there exist \(k\) consecutive composite numbers

$$ t+1,t+2,\ldots,t+k. $$

Suppose that \(n\) is the smallest index such that \(p_n>t+k\). Then

$$ p_n \ge t+k \quad\text{and}\quad p_{n-1},\ldots,p_1 \le t. $$

Hence

$$ p_n \ge t+k \ge k+P_{n-1} \ge 2026+P_{n-1}. \tag{2} $$

From (1) and (2) we obtain a contradiction. Therefore the sequence \( (p_n) \) must be bounded.

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