HMMT 二月 2010 · TEAM1 赛 · 第 4 题

英文解析

4. [ 20 ] Letx ye + e = Ax yxe + ye = B
   2 x 2 yx e + y e = C
   3 x 3 yx e + y e = D
   4 x 4 yx 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 − BCx + y =
   AC − B2
   AE − C2
   x + y =
   AD − BC
   (We have to be careful if AC − B or AD − BC is zero. We’ll deal with that case later.) It is easy to 2
   check that these equations hold by substituting the expressions for A , B , C , D , and E . Setting thesetwo expressions for x + y equal to each other, we get
   AD − BC AE − C2 = ,
   AC − B AD − BC2
   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 + ydown the expressions for A , B , and C gives us that ( x − y ) e = 0, meaning that x = y . If x = y ,
   B E Band x 6 = 0, A and D are also non-zero, and = = x . Since is rational and D is rational, this
   A D Aimplies that E is rational. If x = y = 0, then E = 0 and so is certainly rational.
   We finally must check what happens if AD − BC = 0. Since AD − BC = ( x + y )( AC − B ), either 2
   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,
   C x y 2
   which implies that E = (we know that A 6 = 0 because e and e are both positive). Since A and Care rational, this implies that E is also rational.A
   So, we have shown E to be rational in all cases, as desired.