HMMT 二月 2023 · 冲刺赛 · 第 31 题

HMMT February 2023 — Guts Round — Problem 31

[23] Let 2016 ∏ ( ) 2 3 P = i − i − 1 . i =0 The remainder when P is divided by the prime 2017 is not zero. Compute this remainder. ◦

解析

[23] Let 2016 ∏ ( ) 2 3 P = i − i − 1 . i =0 The remainder when P is divided by the prime 2017 is not zero. Compute this remainder. Proposed by: Srinath Mahankali Answer: 1994 3 Solution 1: Let Q ( x ) = x − x − 1 = ( x − a )( x − b )( x − c ), for a, b, c ∈ F 3 . Then, we can write p 2016 ∏ P = ( i − a )( i − b )( i − c ) . i =0 If we consider each root separately, then 2017 2017 2017 P = − ( a − a )( b − b )( c − c ) . 2017 2017 2017 The key observation is that a , b , c is some nontrivial cycle of a, b, c. This is because by p p p 2017 2017 Frobenius’ identity, ( a + b ) = a + b . So, if P ( x ) = 0 , P ( x ) = 0 . But, x 6 = x , since x 6 ∈ F . p This implies the claim. So, in the end, we wish to compute 2 2 2 ( b − a ) ( c − b ) ( a − c ) . ′ 2 Note that P ( x ) = 3 x − 1 = ( x − b )( x − c ) + ( x − a )( x − b ) + ( x − c )( x − a ) . So it suffices to compute ( ) ( ) 1 1 2 2 2 (1 − 3 a )(1 − 3 b )(1 − 3 c ) = − 27 P √ P − √ . 3 3 ( ) ( ) 2 2 √ √ We can compute this is equal to − 27 1 − 1 + = − 23 . 3 3 3 3 Alternatively, this is the discriminant of P ( x ). We utilize the well-known formula that the discriminant 3 3 2 of x + ax + b is − 4 a − 27 b = − 23 . So, the answer is 1994 . ◦