In my Number Theory Class we found that $7$ was a primitive root of 41 by first finding two integers who have order $5$ and $8$ $modulo 41$ respectively, these being $16$ and $3$. Since $16(3)=7(mod41)$ 7 is a primitive root.
I'm trying to do this for $71$ and so far have that $5$ has order $5(mod71)$.
I'm struggling to find integers whose order $modulo71$ are $2$ and $7$.
$G=\mathbb{Z}/(71\mathbb{Z})^*$ is a cyclic group with order $70=2\cdot 5\cdot 7$. It follows that any $g\in G$ such that $g^{10}\not\equiv 1\pmod{71}$, $g^{14}\not\equiv 1\pmod{71}$ and $g^{35}\not\equiv 1\pmod{71}$ is a generator for $G$. There are $\varphi(70)=1\cdot 4\cdot 6=24$ generators, so a random non-quadratic residue is very likely to be a generator. $71\equiv -1\pmod{8}$ so $2$ is a quadratic residue. $71\equiv -1\pmod{3}$ so $3$ is a quadratic residue too. $4$ is a square, hence a quadratic residue. $71\equiv 1\pmod{5}$, hence $5$ is also a quadratic residue. $6$ too, since it is the product of two quadratic residues ($2$ and $3$). $7$ is the smallest reasonable candidate for a generator. We have $$ 7^{35}\equiv \left(\frac{7}{71}\right)\stackrel{\text{reciprocity}}{\equiv}-\left(\frac{1}{7}\right)\equiv -1\pmod{71}, $$ $$ 7^5\equiv 51 \equiv -20\pmod{71},$$ $$ 7^{10}\equiv (-20)^2 \equiv 45\pmod{71}, $$ $$ 7^{14}\equiv (-20)^3\cdot 7^{-1} \equiv (-20)^3\cdot(-10) \equiv 54\pmod{71} $$ hence $7$ is actually a generator.