Difference between revisions of "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 \pmod(p)$for all$k$. This happens when$p=2, 3$or$5$. When$ (Error compiling LaTeX. ! Missing $inserted.)p=2$then$r$is odd, meaning$ra$,$rb$,$rc$and$rd$are all 1 mod 2, or the sum wouldn't be 2. Any pairwise sum is 2.

When$(Error compiling LaTeX. ! Missing$ inserted.)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$ (Error compiling LaTeX. ! Missing $inserted.)p=5$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.