# Show that there exists no integer $n$ such that $n^3 - n + 3$ divides $n^3 + n^2 + n + 2$

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.

• Hint: $|n^3-n+3|>|n^2+2n-1|$ for $n$ of sufficiently large magnitude. Commented Oct 10, 2022 at 6:56
• 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.
– RDK
Commented Oct 10, 2022 at 6:58
• 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 $n^3−n+3<n^2+2n−1$ Commented Oct 10, 2022 at 7:11
• @BillDubuque I don't see how I can further reduce the degree Commented Oct 10, 2022 at 7:13
• @FadeelKhan Well, you don't really have to go to $\infty$, for all integers, $|n^3-n+3|>|n^2+2n-1|$… Commented Oct 10, 2022 at 8:39

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 $$n=0,\pm1$$).
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.