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Find all natural numbers $n$ such that $$n-2\mid n^2-2$$ I don't know how to find "all natural numbers" Can someone help me prove the statement?

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  • $\begingroup$ By the Factor Theorem $\ n-2\mid f(n)-f(2)\,$ so $\,n-2\mid f(n)\iff n-2\mid f(2),\,$ for $f(x)$ any polynomial with integer coefficient, i.e. $\bmod n-2\!:\ f(n)\equiv f(2)\ \ $ $\endgroup$ – Bill Dubuque Sep 26 '17 at 15:17
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Let us assume then that $n\neq2$. Since $n^2-4=(n-2)(n+2)$, we always have $n-2\mid n^2-4$. If we also have $n-2\mid n^2-2$, then $n-2\mid 2$ and, of course, the reverse is also true. This (I mean, $n-2\mid2$) only happens if $n=1$, $n=3$, or $n=4$.

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  • $\begingroup$ $2$ is not solution, dear friend. $\endgroup$ – Piquito Sep 26 '17 at 13:59
  • $\begingroup$ @Piquito I've edited my answer. Indeed, $0\nmid-2$. $\endgroup$ – José Carlos Santos Sep 26 '17 at 14:02
  • $\begingroup$ How about $n=0$? $\endgroup$ – robjohn Feb 8 '18 at 22:38
  • $\begingroup$ @robjohn As far as I am concerned, it is not a natural number. $\endgroup$ – José Carlos Santos Feb 8 '18 at 22:52
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Hint:

$$\frac{n^2-2}{n-2}=\frac{n^2-2n+2n-2}{n-2}=n+\frac{2n-2}{n-2}=n+\frac{2n-4 +2}{n-2}=n+2+\frac{2}{n-2}$$

Instead of doing these little manipulations step by step you could also use polynomial long division and get the same result.

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$$n-2\mid n^2-2\iff n^2-2=k(n-2)\Rightarrow n=\frac{k\pm\sqrt{k^2-8k+8}}{2}$$ The integer solutions of $k^2-8k+8=x^2$ are $(k,x)=(1,\pm1),(7,\pm1)$ so we get $$n=1,4,3$$

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Hint:

Set $n-2=m\iff n=?$

$n^2=(m+2)^2=m^2+4m+4$

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Assume $n-2\;|\;n^2-2$. Then $n-2\;|\;(n^2-2)-(n-2)=n^2-n=n(n-1)$.

Since $gcd(n-2,n-1)=1$, this implies $n-2\;|\;n$.

Can you go on from there?

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$n-2|n^2-4+2$ so $n-2|(n-2)(n+2)+2$ so $n-2|2$...

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