# 1976 USAMO Problems/Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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

## Solution

Either $a^2=0$ or $a^2>0$. If $a^2=0$, then $b^2=c^2=0$. Symmetry applies for $b$ as well. If $a^2,b^2\neq 0$, then $c^2\neq 0$. Now we look at $a^2\bmod{4}$: $a^2\equiv 0\bmod{4}$: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let $a=2a_1$, $b=2b_1$, and $c=2c_1$. Thus $a_1^2+b_1^2+c_1^2=4a_1^2b_1^2$. 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 $a^2\equiv 0\bmod{4}$. $a^2\equiv 1\bmod{4}$: Since $b^2\neq 0\bmod{4}$, $b^2\equiv 1\bmod{4}$, and $2+c^2\equiv 1\bmod{4}$. But for this to be true, $c^2\equiv 3\bmod{4}$, which is an impossibility. Thus there are no non-zero solutions when $a^2\equiv 1\bmod{4}$.

Thus the only solution is the solution above: $(a,b,c)=0$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 