# 1982 USAMO Problems/Problem 2

## Problem

Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$, $(*)$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$

for $(m,n)=(2,3),(3,2),(2,5)$, or $(5,2)$. Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$.

## Solution 1

Claim Both $m,n$ can not be even.

Proof $x+y+z=0$ , $\implies x=-(y+z)$.

Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$,

by equating cofficient of $y^{m+n}$ on LHS and RHS ,get $\frac{2}{m+n}=\frac{4}{mn}$. $\implies \frac{m}{2} +\frac {n}{2} = \frac{m.n}{2.2}$.

So we have, $\frac{m}{2} | \frac{n}{2}$ and $\frac{n}{2} | \frac{m}{2}$. $\implies m=n=4$.

So we have $S_8=2.(S_4)^2$.

Now since it will true for all real $x,y,z,x+y+z=0$. So choose $x=1,y=-1,z=0$. $S_8=2$ and $S_4=2$ so $S_8 \neq 2 S_4^2$.

This is contradiction !! So, atlest one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even . The cofficient of $y^{m+n-1}$ in $\frac{S_{m+n}}{m+n}$ is $\frac{\binom{m+n}{1} }{m+n} =1$

The cofficient of $y^{m+n-1}$ in $\frac{S_m .S_n}{m.n}$ is $\frac{2}{m}$.

So get $\boxed{m=2}$

Now choose $x=y=\frac1,z=(-2)$.

Since $\frac{S_{n+2}}{2+n}=\frac{S_2}{2}\frac{S_n}{n}$ holds for all real $x,y,z ,x+y+z=0$.

We have , $\frac{2^{n+2}-2}{n+2} = 3 . \frac{2^n-2}{n}$. $\implies \frac{2^{n+1}-1}{n+2} =3.\frac{2^{n-1}-1}{n} \cdots (**)$.

Clearly $(**)$ holds for $n=5,3$ . Even one can say that for $n\ge 6$ , $\text{RHS of (**)} <\text{LHS of (**)}$.

So our answer is $(m,n)=(5,2),(2,5),(3,2),(2,3)$.

-ftheftics

## See Also

 1982 USAMO (Problems • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS