1982 USAMO Problems/Problem 2
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
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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 |
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