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9 | a^2 + b^2 + c^2

Problem:
Prove that if \( 9 | a^2 + b^2 + c^2 \) then 9 divides at least one of \(a^2-b^2\), \(b^2-c^2\) or \( c^2 - a^2 \).

Solution:
Proof by contradiction. Suppose none of them is divisible by 9. Then \(a^2, b^2, c^2 \pmod 9\) are all different. A square number is \( 0, 1, -2, 4 \pmod 9 \). But \( a^2 + b^2 + c^2  \equiv 1-2+4, 0-2+4, 0+1+4, 0+1-2 \pmod 9 \not \equiv 0 \pmod 9\). QED

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