Diophantine Equation y cubed equals x squared plus 2

Theorem
The indeterminate Diophantine equation:
 * $y^3 = x^2 + 2$

has only one solution in the Natural Numbers:


 * $x = 5, y = 3$

Proof
Assume that $x$ is even:

Therefore, the is $2 \paren {2 k^2 + 1} \equiv 2 \pmod 4$

If $y$ is odd, then the will be odd:

and if $y$ is even, then the will be $\equiv 0 \pmod 4$

Therefore, $x$ and $y$ must both be odd since the can never be $\equiv 2 \pmod 4$

Let us rewrite $x$ as $x = y + a$

We now demonstrate that $a$ must be even:

Therefore:

From the, we know that $y$ is odd and therefore $y - 1$ is even and the only even term on the is $2$.

Therefore, our only solution is: $y - 1 = 2 \leadsto y = 3$ and $x = 5$