What $n$ so that $n|m^4+m^3+m^2+k$? What are the conditions of a positive integer $n$ so that for every positive integer $k$, there exist a positive integer $m$ satisfies $n|m^4+m^3+m^2+k$?
For $n=2$, if $k$ is even, then we shall choose an even integer $m$, if $k$ is odd then we shall choose an odd integer $m$.
However, for $n=3$, if $k=1$ then there aren't any $m$ that satisfied the statement, since $m^4+m^3+m^2 = 1$ or $0 $ $(mod $ $3)$ . The same thing happens with $n=4$ $(k=1)$ and $n=5$ $(k=2)$.
I am solving this question by checking when $n$ is an odd prime integer, then I realized that there wouldn't be any $m$ if there existed two different integers $p$ and $q$ $(mod$ $n)$ such that $p^4+p^3+p^2 = q^4+q^3+q^2$ $(mod$ $n)$.
Here I was stuck. Is there a better approach or solution to this problem?
(Sorry, English is my second language)
 A: Definition. For $n\in\Bbb N$, say that $n$ is an apple number if for everey $k\in\Bbb N$ there exists $m\in\Bbb N$ such that $n\mid m^4+m^3+m^2+k$.
Definition. For $n\in\Bbb N$, say that $n$ is a banana number if the map 
$$\begin{align}f_n\colon\quad\Bbb Z/n\Bbb Z&\to \Bbb Z/n\Bbb Z\\x+n\Bbb Z&\mapsto x^4+x^3+x^2+n\Bbb Z\end{align}$$ is injective.
Our task is to determine all apple numbers.
Lemma 1. Every banana number is apple.
Proof. Let $n$ be banana and $k\in\Bbb N$. As $f_n$ is injective, it is also surjective. Hence there exists $m\in\Bbb N$ such that $m^4+m^3+m^2+n\Bbb Z=-k+n\Bbb Z$, i.e., $n\mid m^4+m^3+m^2+k$. $\square$
Lemma 2. Every divisor of an apple number is a banana number.
Proof. Let $n\in\Bbb N$ and $d\mid n$. If $f_d$ is not injective, it is also not surjective. So suppose $a+d\Bbb Z$ is not in its image. If we pick any $k$ with $k\equiv -a\pmod d$, it follows that $m^4+m^3+m^2+k$ is not a multiple of $d$ (nor of $n$), no matter what $m$ is. We conclude that $n$ is not apple. $\square$
Corollary 1. Every apple number is square-free.
Proof. If $d=c^2$ is a perfect square with $c>1$ and $d\mid n$, then $d$ is not banana because $$f_d(c+d\Bbb Z)=0+d\Bbb Z=f_d(0+d\Bbb Z)$$ where $c+d\Bbb Z\ne 0+d\Bbb Z$ (note that $0<c<d$). $\square$
It follows that every apple number is the product of distinct banana primes.
Using the Chinese Remainder Theorem, we see that the converse is also true, i.e.,  

$n$ is apple iff it is the product of distinct banana primes.

Remains the task to determine all banana primes.
By direct computation, we find that $2$ is banana and $3$ is not banana.
Lemma 3. If $p$ is a prime $\equiv 1\pmod 3$ then $p$ is not banana.
Proof. There exists a third root of unity modulo $p$, i.e., a solution (in fact two solutions) of $x^2+x+1\equiv 0\pmod p$. Then
$$ f_p(x+p\Bbb Z)=0+p\Bbb Z=f_p(o+p\Bbb Z)$$ and $p$ is not banana. $\square$
Lemma 4. If $p$ is a prime $\equiv 1\pmod 4$ then $p$ is not banana.
Proof. There exists a fourth root of unity, i.e., $a$ such that $a^4\equiv 1\pmod p$ but $a,a^2,a^3\not\equiv 1\pmod p$.
Then we can find $x$ such that $x\not\equiv 0\pmod p$ and $f_p(x+p\Bbb Z)=f_p(ax+p\Bbb Z)$, which shows that $p$ is not banana.
Indeed, we just need to solve 
$$ x^4+x^3+x^2\equiv a^4x^4+a^3x^3+a^2x^2$$
or equivalently (after rearranging and dividing by $x^2$),
$$ (a^3-1)x+(a^2-1)\equiv 0\pmod p,$$
which has a unique non-zero solution $\bmod p$. $\square$

The situation is still unclear for primes $p\equiv -1\pmod{12}$.
I suspect that $2$ is the only banana prime. In fact, this is readily verified computationally for small $p$ (I tested up to 100000)
If true, this would imply that $n=1$ and $n=2$ are the only apple numbers.
Further evidence for this conjecture is that one can check equations such as
$$ x^4+x^3+x^2\equiv (2x)^4+(2x)^3+(2x)^2\pmod p$$
that has a non-trivial solution iff $15x^2+7x+3\equiv 0$ has, i.e., iff $7^2-4\cdot 3\cdot 15=-131$ is a square $b\bmod p$. For $p\equiv 3\pmod 4$, this is equivalent to $p$ being a square $\bmod{131}$, i.e., this solves the problem for 50% of the remaining prime. Other factors lead to different - and seemingly unrelated - conditions.
