Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
 A: Mod $(2n^3+5)$, you have that $2n^3\equiv-5$. Now suppose $(2n^3+5)$ divides $n^4+n+1$. Then mod $(2n^3+5)$:
$$
\begin{align}
n^4&\equiv-n-1\\
2n^4&\equiv-2n-2\\
n(2n^3)&\equiv-2n-2\\
-5n&\equiv-2n-2\\
-3n&\equiv-2\\
2-3n&\equiv0
\end{align}
$$
So $2n^3+5$ must divide $2-3n$. You can then test values of $n$ to arrive at the inequality.
A: Using the Extended Euclidean Algorithm (which can be applied to $\mathbb{Q}[x]$ as well as $\mathbb{Z}$), we get
$$
\left(36n^2+24n+16\right)\left(n^4+n+1\right)-\left(18n^3+12n^2+8n-27\right)\left(2n^3+5\right)=151
$$
if $2n^3+5\mid n^4+n+1$, we must also have $2n^3+5\mid151$. Since $151$ is prime and there is no integer $n$ so that $2n^3+5=\pm1$ or $2n^3+5=\pm151$, there is no integer $n$ so that $2n^3+5\mid n^4+n+1$.
A: Worth emphasis: it is much easier via mod arithmetic (vs. extended Euclidean algorithm), e.g.
$\!\bmod 2n^3\!+\!5\!:\,\ \color{#c00}{2n^3\equiv -5}\ $ so $\ 0\equiv 2(n^4\!+\!n\!+\!1)\equiv \color{#c00}{2n^3}n\!+\!2n\!+\!2\equiv \color{#0a0}{-3n\!+\!2}$
so $\,\color{#0a0}{3n\equiv 2}\,\Rightarrow\, 0\equiv 3^3(2n^3\!+\!5)\equiv 2(\color{#0a0}{3n)}^3\!+5(27)\equiv 151,\ $ so $\ 2n^3\!+\!5\mid 151$
Remark $ $ Such scalings by the lead coef of the divisor (to eliminate fractions in the division algorithm) are modular forms of the Nonmonic Division Algorithm, e.g. see here for another example using this idea and the Euclidean algorithm.
A: You can use polynomial long division instead:
$$
\require{enclose}
\begin{array}{rll}
    \frac{1}{2}n && \\[-3pt]
   2n^3+5 \enclose{longdiv}{n^4+0n^3+0n^2+n+1}\kern-.2ex \\[-3pt]
      \underline{- (n^4 \quad \quad \quad \quad \ \ + \frac{5}{2}n) \ \ \ } \\[-3pt]
       - \frac{3}{2}n+ 1  \\[-3pt]
  \end{array}
$$
Therefore we have:
$$\frac{n^4+n+1}{2n^3+5} = \frac{1}{2}n + \frac{-\frac{3}{2}n+1}{2n^3+5}$$
and if $2n^3+5 | n^4+n+1$, then $2n^3+5 | 2n^4 + 2n + 2$:
$$\frac{2n^4+2n+2}{2n^3+5} = n + \frac{-3n+2}{2n^3+5}$$
At $n=0$, $-3(0)+2 = 2$ and $2(0)^3+5 = 5$. However, since $n^3$ is monotonically increasing (equivalent to $(n+1)^3-n^3 > 0$), and $-3x+2$ is monotonically decreasing, then $\frac{-3n+2}{2n^3+5} = 1$ does not have solutions in the positive reals. You can repeat this argument for any positive integer $k$.
For negative integers $-k$, we have $-3n+2 = -k(2n^3+5) \Rightarrow 3n-2 = (2k)n^3 - 5k$. At $n=0$, $3(0)-2=-2$ while $2k(0)^3 + 5k = 5k$. Then when $n$ is positive, cubic functions grow much faster than linear solutions, so there are no solutions. The proof of this fact relies on limits.
A: What you've done is entirely correct, I daresay even ingenious, you just haven't taken it all the way to its natural conclusion. Also, it could be written out in a clearer fashion; as is, it reads more like scratchwork, where you've jotted down the key ideas, with the logical connections between them clear to you, but less so to readers, except those who are able to read your mind. Here's one way to express things more fully, including the denouement:
Since 
$$2(n^4+n+1)-n(2n^3+5)=2n^4+2n+2-2n^4-5n=2-3n$$
we have
$$(2n^3+5)\mid(n^4+n+1)\implies(2n^3+5)\mid(2-3n)\implies|2n^3+5|\le|2-3n|$$
Since we are only looking for positive integer solutions, we can remove the absolute values to obtain $2n^3+5\le3n-2$, which is not satisfied by any positive integer $n$.
(Remark: If you want to rule out negative integer solutions as well, you can write $n=-m$ with $m\ge0$ and remove the absolute values to obtain $2m^3-5\le3m+2$, which is satisfied only by $m=1$. But $n=-1$ is not a solution, since $2(-1)^3+5=3$ does not divide $(-1)^4+(-1)+1=1$.)
