缺失的两个整数

设缺失数为 $x,y$，已知其余数为 $z_1,\dots,z_{98}$。

利用总和与平方和：

$$x+y=5050-\sum_{i=1}^{98}z_i,$$ $$x^2+y^2=338350-\sum_{i=1}^{98}z_i^2.$$

由

$$(x+y)^2=x^2+y^2+2xy$$

可解出 $xy$，从而 $x,y$ 是方程 $t^2-(x+y)t+xy=0$ 的两根，可唯一确定。

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Original Explanation

Let the 98 distinct integers be $z_1, z_2, \dots, z_{98}$. If the missing numbers are $x$ and $y$, we know:

1. $$x + y + \sum_{i=1}^{98} z_i = \sum_{n=1}^{100} n = 5050.$$
2. $$x^2 + y^2 + \sum_{i=1}^{98} z_i^2 = \sum_{n=1}^{100} n^2 = \frac{100 \times 101 \times 201}{6} = 338350.$$

So from the sum and the sum of squares, you can solve for $x$ and $y$ uniquely.