We learn typical solution patterns for integer problems.
Pattern ①: Transforming into a Product
Rearrange the equation into the form “A×B=constant,” then examine combinations of divisors.
Example ① (Transforming into a Product)
Find all integer solutions of xy + 2x + 3y = 11.
xy+2x+3y+6 = 17 → (x+3)(y+2) = 17
17 is prime → (x+3, y+2) = (1,17),(17,1),(−1,−17),(−17,−1)
(x,y) = (−2,15),(14,−1),(−4,−19),(−20,−3)
Answer:(x,y) = (−2,15),(14,−1),(−4,−19),(−20,−3)
Find all integer solutions of xy + 2x + 3y = 11.
xy+2x+3y+6 = 17 → (x+3)(y+2) = 17
17 is prime → (x+3, y+2) = (1,17),(17,1),(−1,−17),(−17,−1)
(x,y) = (−2,15),(14,−1),(−4,−19),(−20,−3)
Answer:(x,y) = (−2,15),(14,−1),(−4,−19),(−20,−3)
Pattern ②: Case Analysis by Parity
Example ② (Using Parity)
For positive integers (x,y,z) satisfying x²+y²=z², show that x and y cannot both be odd.
Since the square of an odd number is ≡ 1(mod 4), x²+y²≡2(mod 4). However, z² ≡ 0 or 1(mod 4), so z²≡2 is impossible → contradiction ■
For positive integers (x,y,z) satisfying x²+y²=z², show that x and y cannot both be odd.
Since the square of an odd number is ≡ 1(mod 4), x²+y²≡2(mod 4). However, z² ≡ 0 or 1(mod 4), so z²≡2 is impossible → contradiction ■
Pattern ③: Narrowing Down with mod
Example ③
When p and q are prime and p+q=50, find all pairs (p,q).
If neither is 2, both are odd → their sum is even → a sum of 50 (even) is possible.
However, when p=2, q=48 (not prime). 3: q=47(prime)✓. 5: q=45(×). 7: q=43(✓). 11: q=39(×). 13: q=37(✓). 19: q=31(✓).
Answer:(3,47),(7,43),(13,37),(19,31) and their symmetric pairs
（4 pairs assuming p≤q）
When p and q are prime and p+q=50, find all pairs (p,q).
If neither is 2, both are odd → their sum is even → a sum of 50 (even) is possible.
However, when p=2, q=48 (not prime). 3: q=47(prime)✓. 5: q=45(×). 7: q=43(✓). 11: q=39(×). 13: q=37(✓). 19: q=31(✓).
Answer:(3,47),(7,43),(13,37),(19,31) and their symmetric pairs
（4 pairs assuming p≤q）
Key Points
- Integer solutions → write the equation as product=constant and list divisor pairs
- Parity → consider mod 2 and mod 4
- Prime problems → use the fact that all primes other than 2 are odd
Practice Problems
- Find all integer solutions of xy − x − y = 4.
- Find all positive integer pairs (x,y) satisfying x² − y² = 35.
- Show that n²+n+1 cannot be even when n is an integer.
Answers and Explanations
- (x−1)(y−1)=5. Combinations: (x−1,y−1)=(1,5),(5,1),(−1,−5),(−5,−1) → (x,y)=(2,6),(6,2),(0,−4),(−4,0)
- (x+y)(x−y)=35. x+y>x−y>0（because x,y are positive integers）. 35=35×1=7×5. x+y=35,x−y=1→(18,17). x+y=7,x−y=5→(6,1). (x,y)=(18,17),(6,1)
- n²+n+1=n(n+1)+1. n(n+1) is the product of consecutive integers, so it is even → n(n+1)+1 is odd → it cannot be even ■