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...?)
 A: 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.
A: Because $6$ is not divisible by $4$.
Indeed, if $x$ is even then $y$ is even and $y^2$ is divisible by $4$.
Also, we have that $x^3$ is divisible by $4$.
A: Suppose by contradiction that $x$ is even. Then $x^3$ is a multiple of $8$, so $x^3+6$ would be an even integer not a multiple of $4$. This is impossible since it should be a square.
A: 6 is not a quadratic residue modulo $8$: the quadratic residues are $1,4,0$
If $x=2k$, the equation would give $8k^3+6=y^2$, or in other words, $y^2 \equiv 6 \mod 8$.
There is no such solution since $1,4,0$ are the only elements with square roots.
A: While others showed how to do it better, here's how you can proceed from your calculation:
You derived $y=\sqrt{2(4k^3+3)}=\sqrt{2}\sqrt{4k^3+3}$. Now $4k^3+3$ is odd. Thus $y$ has a factor $\sqrt{2}$ that is not compensated for in the second factor, and therefore cannot be an integer, in contradiction to the assumption.
