HMMT 二月 2015 · 团队赛 · 第 3 题

英文解析

3. [ 15 ] Let z = a + bi be a complex number with integer real and imaginary parts a, b ∈ Z , where i = − 1
   (i.e. z is a Gaussian integer). If p is an odd prime number, show that the real part of z − z is anpinteger divisible by p .
   Answer: N/A Solution 1. We directly compute/expandp p
   Re( z − z ) = Re (( a + bi ) − ( a + bi ))
   [ ( ) ( ) ]
   p pp p − 2 2 p − 4 4 = a − a b + a b − · · · − a.
   2 4
   ( )
   Since is divisible by p for all i = 2 , 4 , 6 , . . . (since 1 ≤ i ≤ p − 1), we havepi
   ( ) ( )
   p pp p − 2 2 p − 4 4 p
   [ a − a b + a b − · · · ] − a ≡ a − a ≡ 0 (mod p )
   2 4
   by Fermat’s little theorem. Thus p divides the real part of z − z .p
   Team
   ( )
   Remark. The fact that is divisible by p (for 1 ≤ i ≤ p − 1) is essentially equivalent to thepip p ppolynomial congruence ( r x + s y ) ≡ r x + s y (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 pz = ( a + bi ) ≡ a + ( bi ) = a ± b i ≡ a ± bi (mod p · Z [ i ]) ,
   u − vwhere we’re using congruence of Gaussian integers (so that u ≡ v (mod p ) if and only if is a
   Gaussian integer). This is equivalent to the simultaneous congruence of the real and imaginary partsppmodulo p , so the real part of z is congruent to a , the real part of z . So indeed p divides the real partpof z − z .