Prove that the equation
$x^{3} + y^{3} + z^{3} = 2002$ has no solutions in integers.

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Solution

We first notice that a cube of an integer is congruent to either $0, 1$ or $- 1$ (mod $9$ ).
Then the sum of two cubes can be congruent to $0, 1, - 1, 2$ or $- 2$ and the sum of three cubes to $0, 1, - 1, 2, - 2, 3$ or $- 3$ .
Since $2002$ is congruent to $4$ (mod $9$ ) the equation has no solution.

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