Difference between revisions of "1982 USAMO Problems/Problem 2"
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== Solution 1 == | == Solution 1 == | ||
− | |||
Claim Both <math>m,n</math> can not be even. | Claim Both <math>m,n</math> can not be even. | ||
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<math>\frac{2}{m+n}=\frac{4}{mn}</math>. | <math>\frac{2}{m+n}=\frac{4}{mn}</math>. | ||
− | <math>\implies \frac{m}{2} +\frac {n}{2} = \frac{m | + | <math>\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}</math>. |
− | So we have, <math>\frac{m}{2} | \frac{n}{2} </math> and <math>\frac{n}{2} | \frac{m}{2}</math>. | + | So we have, <math>\frac{m}{2} \biggm{|} \frac{n}{2} </math> and <math>\frac{n}{2} \biggm{|} \frac{m}{2}</math>. |
<math>\implies m=n=4</math>. | <math>\implies m=n=4</math>. | ||
− | So we have <math>S_8=2 | + | 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>. | Now since it will true for all real <math>x,y,z,x+y+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>. | <math>S_8=2</math> and <math>S_4=2</math> so <math>S_8 \neq 2 S_4^2</math>. | ||
− | This is contradiction | + | 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> |
− | So, | ||
− | The | ||
− | The | + | 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>. |
− | + | Therefore, <math>\boxed{m=2}</math>. | |
− | Now choose <math>x=y=\frac1,z=(-2)</math>. | + | 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 | + | 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 | + | We have <math>\frac{2^{n+2}-2}{n+2} = 3\cdot\frac{2^n-2}{n}</math>. 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 <math>( | + | 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 ( | + | <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>. | So our answer is <math>(m,n)=(5,2),(2,5),(3,2),(2,3)</math>. | ||
− | -ftheftics | + | -ftheftics (edited by integralarefun) |
== See Also == | == See Also == |
Latest revision as of 10:38, 25 March 2023
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
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, at least one of must be odd. WLOG assume
is odd and m is even. The coefficient of
in
is
The coefficient of in
is
.
Therefore, .
Now choose . (sic)
Since holds for all real
such that
.
We have . 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 holds for
.
And one can say that for
,
.
So our answer is .
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.