Difference between revisions of "1976 USAMO Problems/Problem 3"
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*Case 2: 3 is not a divisor of <math>a</math>. | *Case 2: 3 is not a divisor of <math>a</math>. | ||
Thus <math>a^2\equiv b^2\equiv 1\bmod{3}</math>, but for <math>a^2+b^2+c^2</math> to be a quadratic residue, <math>c^2\equiv 1\bmod{3}</math>, and we have that a multiple of 3 equals something that isn't a multiple of 3, which is a contradiction. | Thus <math>a^2\equiv b^2\equiv 1\bmod{3}</math>, but for <math>a^2+b^2+c^2</math> to be a quadratic residue, <math>c^2\equiv 1\bmod{3}</math>, and we have that a multiple of 3 equals something that isn't a multiple of 3, which is a contradiction. | ||
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Thus after both cases, the only solution is the trivial solution stated above. | Thus after both cases, the only solution is the trivial solution stated above. | ||
Revision as of 17:07, 4 October 2008
Problem
Determine all integral solutions of 
.
Solution
We have the trivial solution, 
. Now WLOG, let the variables be positive.
- Case 1:
Thus the RHS is a multiple of 3, and 
and 
are also multiples of 3. Let 
, 
, and 
. Thus 
. Thus the new variables are all multiples of 3, and we continue like this infinitely, and thus there are no solutions with 
.
- Case 2: 3 is not a divisor of
.
.Thus 
, but for 
to be a quadratic residue, 
, and we have that a multiple of 3 equals something that isn't a multiple of 3, which is a contradiction.
Thus after both cases, the only solution is the trivial solution stated above.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
| 1976 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
