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?

$a^4-1 \neq 1+a+a^2+a^3$, but it is true that $a^4-1=(a-1)(a^3+a^2+a+1)=0 \mod p $, and since there are no zero divisiors, either $a=1$, in which case its order cannot be $4$,  or, as you conclude: $(a^3+a^2+a+1)=0$
