Fermat's Little Theorem: Proving an integer exists which satisfies the following criteria mod $5$ This question has been tantalizing me for a day now:

Given integers $a,b,c,d$ such that $d \neq 0$ (mod $5$) and $m$ an integer for which $am^3 + bm^2 +cm+d \equiv 0$ (mod $5$), prove there exists an integer $n$ for which $dn^3+cn^2+bn+a \equiv 0 $ (mod $5$).

My attempt:

It is easy to see that $\gcd(5,m)=1$. We know from Fermat's little theorem that $5 \mid m^4 -1 \Rightarrow 5 \mid (m+1)(m-1)(m^2+1)$. The proof is trivial that $5$ only divides one of these factors. Hence we have $3$ cases:
Case $1 \rightarrow m \equiv 1$ (mod $5$). If this is true then  it follows that $m^2 \equiv 1$ (mod $5$) and $m^3 \equiv 1$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$a+b+c+d \equiv 0 \pmod5$$
$$\Rightarrow d(1)^3+c(1)^2+b(1)+a \equiv 0 \pmod5$$
Where $1$ satisfies the condition.
Case $2 \rightarrow m \equiv -1$ (mod $5$). If this is true then  it follows that $m^2 \equiv 1$ (mod $5$) and $m^3 \equiv -1$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$a-b+c-d \equiv 0 \pmod5$$
$$\Rightarrow d(-1)^3+c(-1)^2+b(-1)+a \equiv 0 \pmod5$$
Where $-1$ satisfies the condition.
Case $3 \rightarrow m^2 \equiv -1$ (mod $5$). If this is true then  it follows that $m \equiv 2$ (mod $5$) and $m^3 \equiv -2$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$2a+b-2c-d \equiv 0 \pmod5$$
From where I am unable to proceed. Can I get a hint?

$m\not\equiv0\bmod5$, since otherwise $d\equiv0\bmod5$, contradicting the given facts. Hence, since $5$ is a prime number, $m$ has an inverse modulo $5$ – let $n$ be this inverse. Multiplying $am^3+bm^2+cm+d\equiv0\bmod5$ by $n^3$, we can easily verify that $n$ satisfies the final congruence.
