I am trying to follow the working of an example of stated algorithm on following link (http://www-math.ucdenver.edu/~wcherowi/courses/m5410/phexam.html).
Let the prime $p = 8101$, and a generator of $Z_{8101}$ be $a = 6$. Find $x$ so that $a^x = 7531 \mod 8101$.
The author states that $c_0,c_1$ can be either $0$ or $1$ and "$7531^{\frac{(p-1)}{2}}= 7531^{4050}= -1$ and as this $= a^{c_0\frac{(p-1)}{2}}$, we have $c_0 = 1$".
However I have calculated $6^{4500} \mod 8101$ as $4334$ which is not congruent to $-1$.
My calculations $\mod 8101$ are as follows;
$$6^3=214$$ $$6^5=-397$$ $$6^{10}=3690$$ $$6^{100}=2117$$ $$6^{1000}=1085$$ $$6^{4500}=4334$$
Where am I going wrong? Am I misunderstanding their working?
Corrections to calculations $\mod 8101$ below;
$$6^3=216$$ $$6^5=7776$$ $$6^{10}=312$$ $$6^{100}=-44$$ $$6^{1000}=3446$$ $$6^{4050}=-1$$