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Singapore Mathematical Olympiad (SMO)

Number theory Difficulty 5.1 AIME, harder Prove it Singapore

Suppose $p$ is a prime number and $x, y, z$ are integers satisfying $0 < x < y < z < p$ . If $x^3, y^3, z^3$ have equal remainders when divided by $p$ , prove that $x^2 + y^2 + z^2$ is divisible by $x + y + z$ .

Solution

Note that $p > 3$ and $p \mid x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ . Since $x, y < p$ , $p \nmid x-y$ . Therefore $p \mid x^2 + xy + y^2$ . Similarly $p \mid x^2 + xz + z^2$ , $p \mid y^2 + yz + z^2$ .

Thus $p \mid (x^2 + xy + y^2) - (y^2 + yz + z^2) = (x-z)(x+y+z)$ and so $p \mid x+y+z$ .

Since $x, y, z < p$ and $p$ is a prime, $x+y+z = p$ or $2p$ .

Since the parity of $x+y+z$ and $x^2+y^2+z^2$ are the same, it suffices to prove that $p \mid x^2+y^2+z^2$ .

Now $p \mid x^2 + xy + y^2 = x(x+y+z) + y^2 - xz$ and so $p \mid y^2 - xz$ . Therefore $p \mid (x^2 + xy + y^2) + (y^2 - xz) = x^2 + y^2 + z^2$ and we are done.

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