HMMT 二月 2015 · 团队赛 · 第 3 题

HMMT February 2015 — Team Round — Problem 3

[ 15 ] Let z = a + bi be a complex number with integer real and imaginary parts a, b ∈ Z , where i = − 1 p (i.e. z is a Gaussian integer). If p is an odd prime number, show that the real part of z − z is an integer divisible by p .

解析

[ 15 ] Let z = a + bi be a complex number with integer real and imaginary parts a, b ∈ Z , where i = − 1 p (i.e. z is a Gaussian integer). If p is an odd prime number, show that the real part of z − z is an integer divisible by p . Answer: N/A Solution 1. We directly compute/expand p p Re( z − z ) = Re (( a + bi ) − ( a + bi )) [ ( ) ( ) ] p p p p − 2 2 p − 4 4 = a − a b + a b − · · · − a. 2 4 ( ) p Since is divisible by p for all i = 2 , 4 , 6 , . . . (since 1 ≤ i ≤ p − 1), we have i ( ) ( ) p p p p − 2 2 p − 4 4 p [ a − a b + a b − · · · ] − a ≡ a − a ≡ 0 (mod p ) 2 4 p by Fermat’s little theorem. Thus p divides the real part of z − z . Team ( ) p Remark. The fact that is divisible by p (for 1 ≤ i ≤ p − 1) is essentially equivalent to the i p p p polynomial congruence ( rX + sY ) ≡ rX + sY (mod p ) (here the coefficients are taken modulo p ), a fundamental result often called that “Frobenius endomorphism”. Solution 2. From the Frobenius endomorphism, p p p p p p z = ( a + bi ) ≡ a + ( bi ) = a ± b i ≡ a ± bi (mod p · Z [ i ]) , u − v where we’re using congruence of Gaussian integers (so that u ≡ v (mod p ) if and only if is a p Gaussian integer). This is equivalent to the simultaneous congruence of the real and imaginary parts p modulo p , so the real part of z is congruent to a , the real part of z . So indeed p divides the real part p of z − z .