We want to solve the equation $y^2=x^3 - 2$ over the integers.

By writing $y^2+2=x^3$ we can factor on as$$(y-i\sqrt{2})(y+i\sqrt{2})=x^3.$$

Using congruences modulo $8$ , one can show that both $x,y$ must be odd, and it can also be shown that $(y-i\sqrt{2})$ and $(y+i\sqrt{2})$ are relatively prime (if it were not the case, any divisor would have even norm, which is not possible).

Therefore, by unique factorization, and using that the only units on are $1,-1$ , we have that each factor must be a cube.

So let us write$$ (y+i\sqrt{2}) = (a+bi\sqrt{2})^3 = (a^3 - 6ab^2) + i(3a^2b-2b^3)\sqrt{2}$$

Then $y=a^3 - 6ab^2$ and $1=3a^2b-2b^3=b(3a^2-2b^2)$ . These two equations imply $b=\pm 1$ and thus $a=\pm 1$ , from where the only possible solutions are $x=3, y=\pm 5$ .

Bibliography

1

Esmonde, Ram Murty; Problems in Algebraic Number Theory. Springer.