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My attempt:

For $n^3 - n + 3$ to divide $n^3 + n^2 + n + 2$, it should also divide $(n^3 + n^2 + n + 2) - (n^3 - n + 3) = n^2 + 2n - 1$. I did this to reduce the degree, but I don't think it helps.

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    $\begingroup$ Hint: $|n^3-n+3|>|n^2+2n-1|$ for $n$ of sufficiently large magnitude. $\endgroup$ Commented Oct 10, 2022 at 6:56
  • $\begingroup$ Well Done... Reducing the degree is the best way to solve this kind of problem. Now, you know that when $n$ increases, $n^3$ increases faster than $n^2$, and at the specific moment, $n^3-n+3$ will be bigger than $n^2+2n-1$, while $n^3-n+3|n^2+2n-1$ so it's contradiction. $\endgroup$
    – RDK
    Commented Oct 10, 2022 at 6:58
  • $\begingroup$ Ok, this works when $n^3−n+3>n^2+2n−1$, but as $n \to -\infty$, $n^3−n+3 \to -\infty$, while $n^2+2n−1 \to \infty$, so there do exist $n$ such that $n^3−n+3<n^2+2n−1$ $\endgroup$ Commented Oct 10, 2022 at 7:11
  • $\begingroup$ @BillDubuque I don't see how I can further reduce the degree $\endgroup$ Commented Oct 10, 2022 at 7:13
  • $\begingroup$ @FadeelKhan Well, you don't really have to go to $\infty$, for all integers, $|n^3-n+3|>|n^2+2n-1|$… $\endgroup$
    – Macavity
    Commented Oct 10, 2022 at 8:39

2 Answers 2

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Here is another way to show this. $n^3-n+3=n(n^2-1)+3=(n-1)n(n+1)+3$, so is obviously divisible by $3$. However $n^3+n^2+n+2$ isn't divisible by $3$ (you need to only check for $n=0,\pm1$).

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If some $n$ works, then the greatest common divisor of $n^3 - n + 3$ and $n^3 + n^2 + n + 2$ must be $|n^3 - n + 3|$. But by the Euclidean algorithm \begin{align*} &\phantom{=} (n^3 - n + 3, n^3 + n^2 + n+ 2) \\ &= (n^3 - n + 3, n^2 + 2n - 1)\\ &= (n^2 + 2n - 1, - 2n^2 + 3)\\ &= (n^2 + 2n - 1, 4n+5)\\ &= (4n^2 + 8n - 4, 4n+5)\\ &\text{[where we note that }4n+5 \text{ is odd hence it cannot be divided by } 4]\\ &= (4n+5, 3n - 4)\\ &= (3n - 4, n + 9)\\ &= (-31, n + 9)\in \{\pm 1, \pm 31\}, \end{align*} thus $|n^3 - n + 3| = (31, n+9)$. So possible cases for $n^3 - n + 3$ are $\pm 1, \pm 31$, or $n^3 - n \in \{-34, -4, -2, 28\}$. Note that $f(x) := x^3 - x, x\in \Bbb R$ is odd and increases on $(-\infty, -1), (1, +\infty)$ respectively, and that $f(-1) = f(1)= f(0) = 0$.

Try values: $f(3) = 24 < 28 < 34, f(4) = 60 > 34 > 28$, so both $-34$ and $28$ cannot be attained. $f(-2) = -f(2) = -6 < -4 $, so $-4, -2$ are also out of range. Therefore no $n$ works.

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