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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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

1982 USAMO (Problems • Resources)
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1982 USAMO Problems/Problem 2

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Solution 1

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