HMMT 二月 2016 · 团队赛 · 第 5 题

英文解析

5. [ 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 pordered 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 xand − 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.3
   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 isnonzero. For any fixed x , there are either 0 or 2 choices for y . Replacing x by − x negates the righthand side, again producing two choices for y since − 1 is a quadratic residue. So the total number ofsolutions ( 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 forthe number of solutions is given in http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/
   026.pdf . 1