Saudi Arabian Mathematical Competitions

Suppose there exists a prime divisor $p \equiv 2 \pmod{3}$ of $x + y + z$. Since $z \equiv -(x + y) \pmod{p}$ we have $2(x^2 + y^2 + xy) \equiv 0 \pmod{p}$. Multiplying by $x - y$ we get $x^3 \equiv y^3 \pmod{3}$ but this yields $x \equiv y \pmod{p}$, because $\gcd(3, p-1) = 1$. Similarly $x \equiv z \pmod{p}$ whence $3x \equiv 0 \pmod{p}$ which means that $x$, $y$ and $z$ are divisible by $p$ which contradicts $\gcd(x, y, z) = 1$. Hence $x + y + z$ have no prime divisors which have a remainder $2$ modulo $3$ so $x + y + z - 2$ is not divisible by $3$.
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