Difference between revisions of "1976 USAMO Problems/Problem 3"
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== Solution == | == Solution == | ||
We have the trivial solution, <math>(a,b,c)=(0,0,0)</math>. Now WLOG, let the variables be positive. | We have the trivial solution, <math>(a,b,c)=(0,0,0)</math>. Now WLOG, let the variables be positive. | ||
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* Case 1: <math>3|a</math> | * Case 1: <math>3|a</math> | ||
Thus the RHS is a multiple of 3, and <math>b</math> and <math>c</math> are also multiples of 3. Let <math>a=3a_1</math>, <math>b=3b_1</math>, and <math>c=3c_1</math>. Thus <math>a_1^2+b_1^2+c_1^2=9a_1^2b_1^2</math>. Thus the new variables are all multiples of 3, and we continue like this infinitely, and thus there are no solutions with <math>3|a</math>. | Thus the RHS is a multiple of 3, and <math>b</math> and <math>c</math> are also multiples of 3. Let <math>a=3a_1</math>, <math>b=3b_1</math>, and <math>c=3c_1</math>. Thus <math>a_1^2+b_1^2+c_1^2=9a_1^2b_1^2</math>. Thus the new variables are all multiples of 3, and we continue like this infinitely, and thus there are no solutions with <math>3|a</math>. | ||
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 | ||
