1976 USAMO Problems/Problem 3

Revision as of 17:06, 4 October 2008 by 1=2 (talk | contribs) (Solution)

Problem

Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$.

Solution

We have the trivial solution, $(a,b,c)=(0,0,0)$. Now WLOG, let the variables be positive.

  • Case 1: $3|a$

Thus the RHS is a multiple of 3, and $b$ and $c$ are also multiples of 3. Let $a=3a_1$, $b=3b_1$, and $c=3c_1$. Thus $a_1^2+b_1^2+c_1^2=9a_1^2b_1^2$. Thus the new variables are all multiples of 3, and we continue like this infinitely, and thus there are no solutions with $3|a$.

  • Case 2: 3 is not a divisor of $a$.

Thus $a^2\equiv b^2\equiv 1\bmod{3}$, but for $a^2+b^2+c^2$ to be a quadratic residue, $c^2\equiv 1\bmod{3}$, 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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions