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>. | ||
− | == Solution == | + | == Solution 1 == |
{{solution}} | {{solution}} | ||
+ | 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.n}{2.2}</math>. | ||
+ | |||
+ | So we have, <math>\frac{m}{2} | \frac{n}{2} </math> and <math>\frac{n}{2} | \frac{m}{2}</math>. | ||
+ | |||
+ | <math>\implies m=n=4</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | <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, atlest one of <math>m,n</math> must be odd. WLOG assume <math>n</math> is odd and m is even . | ||
+ | The cofficient 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 cofficient of <math>y^{m+n-1}</math> in <math>\frac{S_m .S_n}{m.n}</math> is <math>\frac{2}{m}</math>. | ||
+ | |||
+ | So get <math>\boxed{m=2}</math> | ||
+ | |||
+ | Now choose <math>x=y=\frac1,z=(-2)</math>. | ||
+ | |||
+ | 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 ,x+y+z=0</math>. | ||
+ | |||
+ | We have ,<math>\frac{2^{n+2}-2}{n+2} = 3 . \frac{2^n-2}{n}</math>. | ||
+ | |||
+ | <math>\implies \frac{2^{n+1}-1}{n+2} =3.\frac{2^{n-1}-1}{n} \cdots (**)</math>. | ||
+ | |||
+ | Clearly <math>(**)</math> holds for <math>n=5,3</math> . | ||
+ | Even one can say that for <math>n\ge 6</math> , | ||
+ | <math>\text{RHS of (**)} <\text{LHS of (**)}</math>. | ||
+ | |||
+ | |||
+ | So our answer is <math>(m,n)=(5,2),(2,5),(3,2),(2,3)</math>. | ||
+ | |||
+ | -ftheftics | ||
== See Also == | == See Also == |
Revision as of 17:31, 5 March 2020
Problem
Let with real. It is known that if ,
for , or . Determine all other pairs of integers if any, so that holds for all real numbers such that .
Solution 1
This problem needs a solution. If you have a solution for it, please help us out by adding it. Claim Both can not be even.
Proof ,.
Since ,
by equating cofficient of on LHS and RHS ,get
.
.
So we have, and .
.
So we have .
Now since it will true for all real . So choose .
and so .
This is contradiction !! So, atlest one of must be odd. WLOG assume is odd and m is even . The cofficient of in is
The cofficient of in is .
So get
Now choose .
Since holds for all real .
We have ,.
.
Clearly holds for . Even one can say that for , .
So our answer is .
-ftheftics
See Also
1982 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 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.