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.
