A square number and a positive cubic number which differ by six

Question

Is there a square number and a positive cubic number (both positive integers) which differ by six? If not, how do we prove this?

Answer 1

Nice question. Turns out it's not that elementary (?)

The answer is no. Source: http://en.wikipedia.org/wiki/Mordell_equation

Answer 2

Here is a proof that $y^2 = x^3 - 6$ has no solutions, taken from here.

First note modulo $8$ that $y^2$ is one of $0,1, 4$ and $x^3$ is one of $0, 1, 3, 5, 7$. Thus, we must have $y^2 \equiv 1$ and $x^3 \equiv 7$. I.e., $y$ is odd and $x \equiv 7 \pmod 8$. Now write $$ y^2 - 2 = (x - 2)(x^2 + 2x + 4) $$ Since $x - 2 \equiv 5 \pmod 8$, a prime $p \; \mid \; x - 2$ congruent to $3$ or $5$ mod $8$. Then $p \; \mid \; y^2 - 2$, so $2$ is a perfect square modulo $p$. But by a property of the Legendre symbol, letting $p = 8k \pm 3$, we have $$ \left( \frac{2}{p} \right) = (-1)^{\frac{p^2 - 1}{8}} = (-1)^{8k^2 \pm 6k + 1} = -1, $$ which is a contradiction.