1976 USAMO Problems/Problem 3
Problem
Determine all integral solutions of .
Solution
Either or . If , then . Symmetry applies for as well. If , then . Now we look at :
: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let , , and . Thus . Since the LHS is divisible by four, all the variables are divisible by 4, and we must do this over and over again, and from infinite descent, there are no non-zero solutions when .
: Since , , and . But for this to be true, , which is an impossibility. Thus there are no non-zero solutions when .
Thus the only solution is the solution 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 |
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