*Obviously,
one solution is (x, y) = (-1, 0).*

x cannot have any other negative value since y^{2} is
positive.

y^{2} = x^{3} + 1 = (x + 1)(x^{2 }– x
+ 1) =

= (x + 1)^{2} · (x^{2} – x
+ 1)/(x + 1)

To solve the problem we must defactorize and analyze
the last set of binomials:

(x^{2}– x + 1)/(x + 1) =

= (x^{2} + x – 2x + 1)/(x + 1) =

= x – [(2x + 1)/(x + 1)] =

= x – [(2x - 2 + 3)/(x + 1)] =

= x – 2 + 3/(x + 1)

*Now,
0 and 2 are the only integers for which ***3/(x
+ 1)** is an integer. So, the only solutions
for (x, y) other than (**-1**, **0**)
are (**0**, **+/-1**) and
(**2**,** +/-3**).

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