We are given that $ a + b = 1000 $, and $ p \mid a $, $ p \mid b $, so $ p \mid (a + b) = 1000 $. Therefore, $ p $ must be a prime divisor of 1000.

["Understanding Why $ p \mid (a + b) = 1000 $ When $ p \mid a $ and $ p \mid b $", "In number theory and algebra, investigating divisibility conditions often reveals elegant insights. Consider the equation:", "$$\na + b = 1000\n$$", "We’re also given that a prime number $ p $ divides both $ a $ and $ b $:\n$$\np \mid a \quad \ ext{and} \quad p \mid b\n$$", "From basic divisibility, if a prime $ p $ divides both $ a $ and $ b $, then $ p $ must divide their sum:", "$$\np \mid (a + b) = 1000\n$$", "This means $ p $ is a prime divisor of 1000.", "### Why Prime Divisors of 1000?", "Since $ a + b = 1000 $, the prime factors of 1000 directly determine which primes can divide $ a $ and $ b $ when $ p \mid a $ and $ p \mid b $.", "Let’s analyze the number 1000:", "$$\n1000 = 10^3 = (2 \cdot 5)^3 = 2^3 \cdot 5^3\n$$", "Thus, the prime factorization of 1000 contains only the primes $ 2 $ and $ 5 $. Therefore, any prime $ p $ that divides $ a + b = 1000 $ must be either $ 2 $ or $ 5 $.", "This insight is crucial in many problems:\n- It restricts the possible values of $ p $ solely to these two primes.\n- It allows efficient factor checking — only steady divisors of 1000 need to be tested.\n- It highlights the power of divisibility rules in partitioning problems involving sums.", "### Key Takeaways", "- If $ p \mid a $ and $ p \mid b $, then $ p \mid (a + b) $.\n- Here, $ a + b = 1000 $, so $ p \mid 1000 $.\n- The only prime divisors of 1000 are $ 2 $ and $ 5 $.\n- Hence, $ p $ must be $ 2 $, $ 5 $, or potentially $ 1 $ (not prime), so only these two primes qualify.", "### Applications", "This property is widely used in:\n- Modular arithmetic and cryptography\n- Diophantine equation solving\n- Divisibility proofs and number theory exercises", "By focusing on the prime divisors of $ a + b $, we streamline the search for valid divisors and deepen understanding of the structure behind seemingly simple equations.", "---", "Bottom Line:\nWhen $ a + b = 1000 $ and $ p $ divides both $ a $ and $ b $, then $ p $ must divide $ 1000 $, so $ p \in {2, 5} $. This elegant constraint simplifies analysis and highlights the fundamental role of prime factorization in divisibility problems."]