1999 USAMO Problems/Problem 3

Problem

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 $x$ .)

Solution

We see that $\biggl\{\frac{ra+rb+rc+rd}{p}\biggr\}=0$ means that $p|r(a+b+c+d)$ . Now, since $p$ does not divide $r$ and $p$ is prime, their GCD is 1 so $p\mathrel{|}a+b+c+d$ .

Since $\biggl\{\frac{ra}{p}\biggr\}+\biggl\{\frac{rb}{p}\biggr\}+\biggl\{\frac{rc}{p}\biggr\}+\biggl\{\frac{rd}{p}\biggr\}=2$ , then we see that they have to represent mods $\bmod\medspace p$ , and thus, our possible values of $p$ are all such that $k^4 \equiv 1\pmod{p}$ for all $k$ that are relatively prime to $p$ . This happens when $p=3$ or $5$ .

When $p=3$ then $r$ is not divisible by 3, thus two are $1$ , and the other two are $2$ . Thus, four pairwise sums sum to 3.

When $p=5$ then $r$ is not divisible by 5 so $a, b, c, d$ are $1, 2, 3,$ and $4$ , so two pairwise sums sum to 5.

All three possible cases work so we are done.

(This solution makes absolutely no sense. Why is $k^4\equiv 1$ ? And how do we know that only $3$ and $5$ work!?) why can't you have a,b,c,d = 2,2,2,4 p = 5 r=1? it doesn't say distinct integers

1999 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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1999 USAMO Problems/Problem 3

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