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Difference between revisions of "1982 USAMO Problems/Problem 2"

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for <math>(m,n)=(2,3),(3,2),(2,5)</math>, or <math>(5,2)</math>. Determine ''all'' other pairs of integers <math>(m,n)</math> if any, so that <math>(*)</math> holds for all real numbers <math>x,y,z</math> such that <math>x+y+z=0</math>.
 
for <math>(m,n)=(2,3),(3,2),(2,5)</math>, or <math>(5,2)</math>. Determine ''all'' other pairs of integers <math>(m,n)</math> if any, so that <math>(*)</math> holds for all real numbers <math>x,y,z</math> such that <math>x+y+z=0</math>.
  
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== Solution ==
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== Solution 1 ==
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{{solution}}
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Claim Both <math>m,n</math> can not be even.
 +
 
 +
Proof
 +
<math>x+y+z=0</math> ,<math>\implies x=-(y+z)</math>.
 +
 
 +
Since <math>\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}</math>,
 +
 
 +
by equating cofficient of <math>y^{m+n}</math> on LHS and RHS ,get
 +
 
 +
<math>\frac{2}{m+n}=\frac{4}{mn}</math>.
 +
 
 +
<math>\implies  \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}</math>.
 +
 
 +
So we have, <math>\frac{m}{2} \biggm{|} \frac{n}{2} </math> and <math>\frac{n}{2} \biggm{|} \frac{m}{2}</math>.
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<math>\implies m=n=4</math>.
 +
 
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So we have <math>S_8=2(S_4)^2</math>.
 +
 
 +
Now since it will true for all real <math>x,y,z,x+y+z=0</math>.
 +
So choose <math>x=1,y=-1,z=0</math>.
 +
 
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<math>S_8=2</math> and <math>S_4=2</math> so <math>S_8 \neq 2 S_4^2</math>.
 +
 
 +
This is contradiction. So, at least one of <math>m,n</math> must be odd. WLOG assume <math>n</math> is odd and m is even. The coefficient of <math>y^{m+n-1}</math> in <math>\frac{S_{m+n}}{m+n}</math> is <math> \frac{\binom{m+n}{1} }{m+n} =1</math>
 +
 
 +
The coefficient of <math>y^{m+n-1}</math> in <math>\frac{S_m\cdot S_n}{m\cdot n}</math> is <math>\frac{2}{m}</math>.
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Therefore, <math>\boxed{m=2}</math>.
 +
 
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Now choose <math>x=y=\frac1,z=(-2)</math>. (sic)
 +
 
 +
Since <math>\frac{S_{n+2}}{2+n}=\frac{S_2}{2}\frac{S_n}{n}</math> holds for all real <math>x,y,z</math> such that <math>x+y+z=0</math>.
 +
 
 +
We have <math>\frac{2^{n+2}-2}{n+2} = 3\cdot\frac{2^n-2}{n}</math>. Therefore,
 +
 
 +
\begin{equation*}
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\label{eq:l2}
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\frac{2^{n+1}-1}{n+2} =3\cdot\frac{2^{n-1}-1}{n}\ldots
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\tag{**}
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\end{equation*}
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Clearly <math>(\ref{eq:l2})</math> holds for <math>n\in\{5,3\}</math>.
 +
And one can say that for <math>n\ge 6</math>,
 +
<math>\text{RHS of (\ref{eq:l2})}<\text{LHS of (\ref{eq:l2})}</math>.
 +
 
 +
 
 +
So our answer is <math>(m,n)=(5,2),(2,5),(3,2),(2,3)</math>.
 +
 
 +
-ftheftics (edited by integralarefun)
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:38, 25 March 2023

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

$\frac{2}{m+n}=\frac{4}{mn}$.

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

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

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