Find all integer solutions of the equation: n³ + 1000 = m³?

Answer 1

The only integer solutions of this are:

0³ + 10³ = 10³
and
(-10)³ + 10³ = 0³

Answer 2

The only integer solutions have m = 0 or n = 0.
For if we change n to -n and rewrite the
equation, it becomes
n³ + m³ = 10³,
which has no nontrivial integer solutions,
by Fermat's last theorem for cubes.

Answer 3

Solutions to n^3 +10^3 = m^3 require that n or m = 0
as Euler proved that x^3+y^3 = z^3 only has trivial solutions.

Answer 4

use that
(m-n) (m^2+nm+n^2) =1000
Now since 1000 has a finite number of factorizations, you have a finite number of cases. You can find a smaller number of cases proceeding in the following way:

Since 2 divides 1000 then
2 divides either m-n or m^2+nm +n^2
Now take 4 cases:
n odd or even, m odd or even
divide by a power of 2 in the equation etc

Answer 5

n^3 + 1000 can be factored into (n+10)(n^2 - 10n + 100), but I don't know if that will help you at all.