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Quadratics and divisibility

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

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Seems identical to math.stackexchange.com/questions/233354/… –  dinoboy Nov 13 '12 at 7:29
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marked as duplicate by Andres Caicedo, Cameron Buie, TMM, Old John, Grigory M Nov 13 '12 at 12:04

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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 $p$.

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