1999 USAMO 第 3 题

Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that

$$\left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\} + \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2$$

for any integer $r$ not divisible by $p$. Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$. (Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

令 $p > 2$ 为素数，并令 $a,b,c,d$ 为不可被 $p$ 整除的整数，这样

$$\left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\} + \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2$$

对于任何不能被 $p$ 整除的整数 $r$。证明数字 $a+b$、$a+c$、$a+d$、$b+c$、$b+d$、$c+d$ 中至少有两个可被 $p$ 整除。 (注：$\{x\} = x - \lfloor x \rfloor$ 表示 $x$ 的小数部分。)