For any odd integer $x,y$, $(x^2+2) \nmid (y^2+4)$ 
Possible Duplicate:
Quadratics and divisibility 

Prove that for any odd integers $x$ and $y$, we have $(x^2+2) \nmid (y^2+4)$

If $x$ is odd, then $x^2+2\equiv 3\pmod{4}$, so $x^2+2$ must have a prime divisor $p$ of the form $4k+3$. 
If $x^2+2$ divides $y^2+4$, then $y^2+4\equiv 0 \pmod{p}$, or equivalently $y^2\equiv -4\pmod{p}$. Let $w$ be the inverse of $2$ modulo $p$. Then $w^2y^2\equiv -4w^2\equiv -1\pmod{p}$, and therefore
$$(wy)^2\equiv -1\pmod{p}.$$
This is impossible, since that would mean that $-1$ is a quadratic residue of $p$. But it is a standard theorem that if $p$ is of the shape $4k+3$, then $-1$ is not a quadratic residue of $p$.  
