1976 USAMO Problems/Problem 3

Problem

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

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, $(a,b,c)=(0,0,0)$ . 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 indefinitely, 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.

1976 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1976 USAMO Problems/Problem 3

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