Difference between revisions of "1976 USAMO Problems/Problem 3"
m (→Solution) |
m (→Solution) |
||
Line 9: | Line 9: | ||
* 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 | + | 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 indefinitely, and thus there are no solutions with <math>3|a</math>. |
*Case 2: 3 is not a divisor of <math>a</math>. | *Case 2: 3 is not a divisor of <math>a</math>. |
Revision as of 17:07, 4 October 2008
Problem
Determine all integral solutions of .
Solution
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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 indefinitely, 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.
See also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |