HMMT 二月 2007 · 代数 · 第 9 题

英文解析

9. [ 7 ] The complex numbers α , α , α , and α are the four distinct roots of the equation x + 2 x + 2 = 0.
   1 2 3 4
   Determine the unordered set
   { α α + α α , α α + α α , α α + α α } .
   1 2 3 4 1 3 2 4 1 4 2 3
   √
   Answer: { 1 ± 5 , − 2 } . Employing the elementary symmetric polynomials ( s = α + α + α + α =
   1 1 2 3 4 − 2, s = α α + α α + α α + α α + α α + α α = 0, s = α α α + α α α + α α α + α α α = 0,
   2 1 2 1 3 1 4 2 3 2 4 3 4 3 1 2 3 2 3 4 3 4 1 4 1 2
   and s = α α α α = 2) we consider the polynomial
   4 1 2 3 4
   P ( x ) = ( x − ( α α + α α ))( x − ( α α + α α ))( x − ( α α + α α ))
   1 2 3 4 1 3 2 4 1 4 2 3
   Because P is symmetric with respect to α , α , α , α , we can express the coefficients of its expanded
   1 2 3 4
   form in terms of the elementary symmetric polynomials. We compute
   3 2 2 2
   P ( x ) = x − s x + ( s s − 4 s ) x + ( − s − s s + s s )
   2 3 1 4 4 4 2
   3 1 = x − 8 x − 83 = ( x + 2)( x − 2 x − 4)2
   √ √
   The roots of P ( x ) are − 2 and 1 ± 5, so the answer is { 1 ± 5 , − 2 } .
   Remarks. It is easy to find the coefficients of x and x by expansion, and the constant term can be 2
   computed without the complete expansion and decomposition of ( α α + α α )( α α + α α )( α α +
   1 2 3 4 1 3 2 4 1 4
   6 2
   α α ) by noting that the only nonzero 6 th degree expressions in s , s , s , and s are s and s s . The
   2 3 1 2 3 4 4
   1 1
   general polynomial P constructed here is called the cubic resolvent and arises in Galois theory.
   2007 20062