Functions with modular arithmetic

For all primes $p$ and integers $x$ the little theorem of Fermat applies: $x^{p} = x \quad (\text{mod } p)$ In our case, the modulus $p = 5$ is also a prime, so we can apply the theorem of Fermat to our prime field. Therefore, in $\mathbb{Z} / 5 \mathbb{Z}$ applies for $x \not= 0$ $\begin{aligned} & & x^{5} &= x \\ \Rightarrow & & x^3 \cdot x &= 1 \\ \Rightarrow & & x^{-1} &= x^{3} \end{aligned}$ In particular, $\begin{aligned} 3^{-1} &= 3^3 = 3 \cdot (9 \text{ mod } 5) = 3 \cdot 4 = (12 \text{ mod } 5) = 2 \\ -1 &= (-1 + 5 ) = 4 \end{aligned}$ Therefore, we can rewrite the function $g(x)$ $\begin{aligned} g(x) &= 4 \cdot x^{5} \cdot x^{2}+ 3^{-1} \cdot x^5 \\ &= (-1) \cdot x^3 + 2 \cdot x \\ &= f(x) \end{aligned}$ Thus, both functions are identical.