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)
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.,
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.