<s>[INST] The order of a modulo p. 
If $p$is prime and $\operatorname{ord}_p(a)=4$, then $1+a+a^2+a^3≡ 0\bmod p$, where $\operatorname{ord}_p(a)$ is the order of $a$ modulo $p$.

I think it is true statement 
$\operatorname{ord}_p(a)=4$, then $a^4≡1\bmod p$ so $a^4-1≡0\bmod p$
since $a^4-1=1+a+a^2+a^3$, then $a^4-1=1+a+a^2+a^3\equiv0\bmod p$
is that correct please?
 [/INST] You can also find this directly since $4$ is conveniently small.
Since $p$ is prime, there are two roots of $1$: $\{1,-1\}$. Since $(a^2)^2\equiv 1$ but $a^2\not\equiv 1 \bmod p$ we must have $a^2\equiv \color{red}{-1} \bmod p$. Then $a^3\equiv a^2\cdot a \equiv \color{blue}{-a} \bmod p$.
Thus $1+a+a^2+a^3\equiv 1+a+ \color{red}{-1}+ \color{blue}{-a} \equiv 0\bmod p$.
</s>