1982 USAMO Problems/Problem 2


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


$\implies  \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$.

So we have, $\frac{m}{2} \biggm{|} \frac{n}{2}$ and $\frac{n}{2} \biggm{|} \frac{m}{2}$.

$\implies m=n=4$.

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, at least one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even. The coefficient of $y^{m+n-1}$ in $\frac{S_{m+n}}{m+n}$ is $\frac{\binom{m+n}{1} }{m+n} =1$

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

Therefore, $\boxed{m=2}$.

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

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

We have $\frac{2^{n+2}-2}{n+2} = 3\cdot\frac{2^n-2}{n}$. Therefore,

\begin{equation*} \label{eq:l2} \frac{2^{n+1}-1}{n+2} =3\cdot\frac{2^{n-1}-1}{n}\ldots \tag{**} \end{equation*}

Clearly $(\ref{eq:l2})$ holds for $n\in\{5,3\}$. And one can say that for $n\ge 6$, $\text{RHS of (\ref{eq:l2})}<\text{LHS of (\ref{eq:l2})}$.

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

-ftheftics (edited by integralarefun)

See Also

1982 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5
All USAMO Problems and Solutions

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