Finding perfect cubes.

Since $a^3 \equiv -1, 0,1 \pmod {9}$ for all positive integers $a$ ,

$m^3 + n^3 + 4 \equiv a^3 \pmod 9$

Case 1

$m^3 \equiv -1$ and $n^3 \equiv -1$ .

$m^3 + n^3 + 4 \equiv 2 \pmod {9}$

Case 2

$m^3 \equiv -1$ and $n^3 \equiv 0$ .

$m^3 + n^3 + 4 \equiv 3 \pmod {9}$

Case 3

$m^3 \equiv -1$ and $n^3 \equiv 1$ .

$m^3 + n^3 + 4 \equiv 4 \pmod {9}$

Case 4

$m^3 \equiv 0$ and $n^3 \equiv 0$ .

$m^3 + n^3 + 4 \equiv 4 \pmod {9}$

Case 5

$m^3 \equiv 0$ and $n^3 \equiv 1$ .

$m^3 + n^3 + 4 \equiv 5 \pmod {9}$

Case 6

$m^3 \equiv 1$ and $n^3 \equiv 1$ .

$m^3 + n^3 + 4 \equiv 6 \pmod {9}$

Since none of the final congruences equal to -1, 0 and 1, it is not possible for $m^3 + n^3 + 4$ is a perfect cube.

prove by mod 7