Multiply through by $a^2$ to obtain $$a^4+a^3+a^2+a+1=0$$Multiply by $(a-1)$ to get $$a^5-1=0; a^5=1$$ since $a\neq 1$
Now we can consider the exponents mod $5$.
If $m$ is a multiple of $5$ we have one case. If it is not a multiple of $5$ we have another.
In the first case all the terms are equal to $1$, giving the answer $4$.
In the second case the exponents $m, 2m, -m, -2m$ are all distinct and non-zero mod $5$ [if, say $m\equiv -2m$, we'd have $5|3m$, whence $5|m$ contradiction] and are therefore equivalent in some order to $1,2,3,4$
Then $a^4+a^3+a^2+a=-1$ from the first equation of this answer.