2002年复旦大学自主招生有这样一道题:
证明方程x3-2y3=1的任一组整数解(x,y)(y≠0)都满足|x/y-3√2|＜4/|y3|.
证明:1=x3-2y3=(x-3√2y)(x2+3√2xy+3√4y2),由x2+3√2xy+3√4y2＞x2+3√2xy+1/4(3√2y)2=(x+1/23√2y)2＞0得x-3√2y＞0,下面证明x2+3√2xy+3√4y2＞y2/4.此式可化为(x/y)2+3√2x/y+3√4-1/4＞0,因为△=(3√2)2-43√4+1≈-3.76＜0,所以x2+3√2xy+3√4y2＞y2/4,得证.
1=x3-2y3=(x-3√2y)(x2+3√2xy+3√4y2)＞y2/4(x-3√2y),即|x/y-3√2|＜4/|y3|.
能否缩小不等式|x/y-3√2|＜4/|y3|中的系数4呢?
