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 $\{\frac{ra+rb+rc+rd}{p}\}=0$ means that $p|r(a+b+c+d)$. Now, since $p$ does nto divide $r$ and $p$ is prime, their GCD is 1 so $p|a+b+c+d$.

Since $\{ \frac{ra}{p} \}+\{ \frac{rb}{p} \}+\{ \frac{rc}{p} \}+\{ \frac{rd}{p} \} =2$, then we see that they have to represent mods $\mod p$, and thus, our possible values of $p$ are all such that $k^4 \equiv 1 \mod(p)$ for all $k$. 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 does $k^4\equiv 1$??)

See Also

1999 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAMO Problems and Solutions

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