Order of $2^{36} \pmod{107}$ What is the order of $2^{36}\pmod{ 107}$? 
My current thought is $$2^{106} \equiv 1 \pmod{107}$$ according to Euler's Theorem. However, I don't know how to proceed from here or maybe my approach is wrong from the beginning?

Since $107-1=2\cdot 53$, every quadratic non-residue, with the only exception of $-1$, is a generator of $\mathbb{Z}/(107\mathbb{Z})^*$. $107$ is a prime of the form $8k+3$, hence $\left(\frac{2}{107}\right)=-1$ and $2$ has order $106$ in $\mathbb{Z}/(107\mathbb{Z})^*$. We have $\gcd(36,106)=2$, hence the order of $2^{36}$ in $\mathbb{Z}/(107\mathbb{Z})^*$ is $\frac{106}{2}=\color{red}{53}$.
