Given $a, b, c, d, m$ in $\mathbb{Z}$ such that $5|(am^3 + bm^2 + cm + d)$ and $5 \not| d$ , prove that there exists an integer $n$ such that $5\mid(dn^3 + cn^2 + bn + a)$
I've spent about two hours trying to solve this problem but haven't been able to do so. Here's the furthest I've gotten:
$5\nmid d$ so $5\nmid m$ as well.
Case I: $5\mid a$ ... set $n = 0$
Case II: $5\nmid a$
$(am^3 + bm^2 + cm + d) \equiv 0 \pmod 5$
Then multiply the equation by $(m^3)^{-1}$ to make the $a$ coefficient 1, like so:
$(d (m^3)^{-1} + c m (m^3)^{-1} + b m^2 (m^3)^{-1} + a) \equiv 0 \pmod 5$
Now the question remains... does there exist an integer $n$ such that:
$n \equiv m^2 (m^3)^{-1}$
$n^2 \equiv m (m^3)^{-1}$
$n^3 \equiv (m^3)^{-1}$
You are basically done. Choose $n=m^{-1}$ in $Z_5$. The existence of the multiplicative inverse of $m$ is guaranteed since $5$ is prime, which implies $Z_5$ is a field.