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The statement being $$n^2 + 2 \mid 2014n + 2$$

The answer is $n = -2, 0, 1, 2014$. Don't know how to arrive at this answer without using comp sci. (Using the compsci answer, we can restrict the domain to integers in the range -2500 to 2500 since we require $n^2 + 2 < 2014n+2$)

It's not hard to see (nor come up with) the ones that do work. Proving that others don't, though..

MT_
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    The general case $n^2+2|an+2$ would be especially interesting... – lemon Aug 15 '14 at 22:06
  • @ozo I agree. Is the observation that only a few very small cases and $n=a$ work apply to all of the $a$? – MT_ Aug 15 '14 at 22:10
  • 481 -- -32, -14, -11, -5, -3, -2, 0, 1, 4, 6, 30, 481 xxxxxx 1642 -- -41, -8, -5, -2, 0, 1, 3, 4, 10, 40, 1642 xxxxxx 1625 -- -58, -7, -4, -3, -1, 0, 2, 5, 8, 56, 1625 xxxxxx 707 -- -22, -19, -7, -4, -1, 0, 2, 3, 5, 14, 707 xxxxxx – Ed Pegg Aug 15 '14 at 22:29

1 Answers1

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Hint $\,\ n^2\!+2\mid 2\!+\!an\,\Rightarrow\,n^2\!+2\mid (2\!+\!an)(2\!-\!an)+a^2(n^2\!+2) \,=\, 4+2a^2\,$

Remark $\ $ Intuitively we took the "norm" of a quadratic number, i.e.

$\qquad\qquad\begin{eqnarray} {\rm mod}\ n^2\!+2\!:\,\ n\equiv \sqrt{-2}\ \ \ {\rm so}\ \ \ 0 &\equiv& \alpha \ =\ \ \, 2 + a\sqrt{-2}\,\equiv\, 2+a\,n \\ \Rightarrow\ \ 0&\equiv& \alpha\bar \alpha = (2+a\sqrt{-2})(2-a\sqrt{-2})\, =\, 4 + 2 a^2\end{eqnarray}$

the norm is one standard way to get a simpler multiple, by mapping "irrational" obects into "rational" objects - as we did above. See here for another example.

Alternatively, we constructed a multiple of $\,n^2\!+2\,$ free of $\,n\,$ by eliminating $\,n\,$ from the multiples $\,2\!+\!an\,$ and $\,n^2\!+2,\,$ by taking a suitable linear combination (mechanized via resultants or Bezout).

Bill Dubuque
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  • I realize I'm supposed to have an "aha!" moment right now, but it's not coming. Mind giving me a bit more explanation? I think that this is supposed to show that $n^2 + 2 \mid an+2 <=> n^2 + 2 \mid 2(a^2 + 2)$, but I don't see the steps here. – MT_ Aug 15 '14 at 22:55
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    @MichaelT I added some remarks on the intuition behind the derivation. – Bill Dubuque Aug 15 '14 at 23:19