Use Fermat's little theorem to find the roots of $2x^{219}+3x^{74}+2x^{57}+3x^{44}$ in $\mathbb{Z}_5$

Question

The solution says $2x^{219}+3x^{74}+2x^{57}+3x^{44}=_52x^3+3x^2+2x^1+3x^0$ but I don't see how they arrived at that, even with Fermat's theorem

Answer 1

$5$ is prime, thus by Fermat's theorem $a^4 \equiv 1 (\text{mod } 5)$ for all $a \in \mathbb{Z}_5$, such that $a \neq 0$. Therefore you can reduce any power of $x$ to its remainder after the division by $4$.

For the first coefficient we have then $$ 2x^{219} = 2\cdot (x^4)^{54}\cdot x^3 = 2\cdot 1 \cdot x^3. $$

And as pointed out by André Nicolas, solving the case $a = 0$ aside we can easily see, that it is also a root.

Answer 2

$\forall$ prime $P$, and $\forall$ $a\in \mathbb Z$, $a<P$, Fermat's theorem says that $a^{(P-1)} \equiv 1 (mod P)$ Here $5$ is Prime and hence each power can be reduced to a power less than $5$ because obviously the solution belongs to $\mathbb Z_5$