HMMT 二月 2016 · 团队赛 · 第 5 题
HMMT February 2016 — Team Round — Problem 5
题目详情
- [ 35 ] Find all prime numbers p such that y = x + 4 x has exactly p solutions in integers modulo p . In other words, determine all prime numbers p with the following property: there exist exactly p ordered pairs of integers ( x, y ) such that x, y ∈ { 0 , 1 , . . . , p − 1 } and 2 3 p divides y − x − 4 x. 1
解析
- [ 35 ] Find all prime numbers p such that y = x + 4 x has exactly p solutions in integers modulo p . In other words, determine all prime numbers p with the following property: there exist exactly p ordered pairs of integers ( x, y ) such that x, y ∈ { 0 , 1 , . . . , p − 1 } and 2 3 p divides y − x − 4 x. Proposed by: Evan Chen Answer: p = 2 and p ≡ 3 (mod 4) Clearly p = 2 works with solutions (0 , 0) and (1 , 1) and not (0 , 1) or (1 , 0). 3 3 If p ≡ 3 (mod 4) then − 1 is not a quadratic residue, so for x + 4 x 6 = 0, exactly one of x + 4 x 3 and − x − 4 x is a square and gives two solutions (for positive and negative y ), so there’s exactly two 3 solutions for each such pair { x, − x } . If x is such that x + 4 x = 0, there’s exactly one solution. If p ≡ 1 (mod 4), let i be a square root of − 1 (mod p ). The right hand side factors as x ( x + 2 i )( x − 2 i ). For x = 0 , 2 i, − 2 i this is zero, there is one choice of y , namely zero. Otherwise, the right hand side is nonzero. For any fixed x , there are either 0 or 2 choices for y . Replacing x by − x negates the right hand side, again producing two choices for y since − 1 is a quadratic residue. So the total number of solutions ( x, y ) is 3 (mod 4), and thus there cannot be exactly p solutions. Remark : This is a conductor 36 elliptic curve with complex multiplication, and the exact formula for the number of solutions is given in http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/ 026.pdf . 1