HMMT 二月 2010 · TEAM1 赛 · 第 4 题
HMMT February 2010 — TEAM1 Round — Problem 4
题目详情
- [ 20 ] Let x y e + e = A x y xe + ye = B 2 x 2 y x e + y e = C 3 x 3 y x e + y e = D 4 x 4 y x e + y e = E. Prove that if A , B , C , and D are all rational, then so is E .
解析
- [ 20 ] Let x y e + e = A x y xe + ye = B 2 x 2 y x e + y e = C 3 x 3 y x e + y e = D 4 x 4 y x e + y e = E. Prove that if A , B , C , and D are all rational, then so is E . Solution: We can express x + y in two ways: AD − BC x + y = 2 AC − B 2 AE − C x + y = AD − BC 2 (We have to be careful if AC − B or AD − BC is zero. We’ll deal with that case later.) It is easy to check that these equations hold by substituting the expressions for A , B , C , D , and E . Setting these two expressions for x + y equal to each other, we get 2 AD − BC AE − C = , 2 AC − B AD − BC which we can easily solve for E as a rational function of A , B , C , and D . Therefore if A , B , C , and D are all rational, then E will be rational as well. 2 2 Now, we have to check what happens if AC − B = 0 or AD − BC = 0. If AC − B = 0, then writing 2 x + y down the expressions for A , B , and C gives us that ( x − y ) e = 0, meaning that x = y . If x = y , B E B and x 6 = 0, A and D are also non-zero, and = = x . Since is rational and D is rational, this A D A implies that E is rational. If x = y = 0, then E = 0 and so is certainly rational. 2 We finally must check what happens if AD − BC = 0. Since AD − BC = ( x + y )( AC − B ), either 2 2 AC − B = 0 (a case we have already dealt with), or x + y = 0. But if x + y = 0 then AE − C = 0, 2 C x y which implies that E = (we know that A 6 = 0 because e and e are both positive). Since A and C A are rational, this implies that E is also rational. So, we have shown E to be rational in all cases, as desired.