1976 USAMO Problems/Problem 3
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 |