Showing something is even Show that if $x$ and $y$ are integers and $x^3+6=y^2$, then $x$ is odd. 
I tried to do it by contradiction, with something like :
"Suppose for contradiction $x$ is even, then $x=2k$, and $x^3+6=y^2=8k^3+6=2(4k^3+3)$ which means $y=\sqrt{2(4k^3+3)}=\sqrt{2}\sqrt{4k^3+3}$ but I don't know how to conclude from there that $y$ is not gonna be an integer or some kind of contradiction...
Doing it the other way, with something like y is odd or y is even, I came to conclusion like $x$ is even... ($y$ is even, then $y^2$ is even, then $x^3+6$ is even, then $x^3$ is even, then $x$ is even, I clearly don't know what is wrong with that either...?)

suppose that $x$ is even, $x=2z$, $x^3+6=8z^3+6=y^2$ if $y$ is odd impossible since$y^2$ is odd and $8z^3+6$ is even, if $y$ is even, then $y=2a$ you have $8z^3+6=4a^2$, this implies that $4$ divides $6$ contradiction.
