HMMT 二月 2007 · 代数 · 第 9 题
HMMT February 2007 — Algebra — Problem 9
题目详情
- [ 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 2007 2006
解析
- [ 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 3 = x − 8 x − 8 2 = ( x + 2)( x − 2 x − 4) √ √ The roots of P ( x ) are − 2 and 1 ± 5, so the answer is { 1 ± 5 , − 2 } . 2 Remarks. It is easy to find the coefficients of x and x by expansion, and the constant term can be 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 6th 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. 2 2007 2006