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?

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.
