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
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,
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 |
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