Summary of Properties of Integers ③
- Prime numbers: Test divisibility by primes up to √n. p|ab → p|a or p|b
- Proof by contradiction: Negate the conclusion → contradiction. √p (p prime) is irrational
- Fermat's Little Theorem: ap−1≡1(mod p)（p: prime, p∤a）
- Comprehensive integer problems: products of consecutive integers, transforming into products, and casework using mod
Comprehensive Practice Problems
- Find p such that p, p+2, and p+4 are all prime（p≥2）.
- Find the remainder when 21000 is divided by 13.
- Find the integer solutions among real pairs (x,y) satisfying x² + y = 2024 and y² + x = 2024 simultaneously.
Answers and Explanations
- When p=3, 3,5,7（all prime ✓）. For p≥5, one of p, p+2, or p+4 is a multiple of 3 (three consecutive odd numbers always include a multiple of 3) → it cannot be prime. p=3
- 212≡1(mod 13)（Fermat: p=13,p−1=12）. 1000=12×83+4 → 21000≡2⁴=16≡3(mod 13) → remainder 3
- Subtract the two equations: x²−y²−(x−y)=0 → (x−y)(x+y−1)=0。(i)x=y → x²+x=2024 → no solution (discriminant 1+4×4×2024<0 …in fact x²+x−2024=0 → x=(−1±√8097)/2 is irrational). (ii)x+y=1 → y=1−x → x²+(1−x)=2024 → x²−x−2023=0 → discriminant=1+4×2023=8093（not a perfect square）→ no integer solutions. No integer solutions