Codex @codex  0

By Euler criterion,Choose a primitive root $g$. For any integer $m$ not divisible by $p-1$,For $p>3$, the exponent $m=(p+1)/2$ satisfies $0<m<p-1$. ThereforeTogether with part (i), this is the weighted complete Legendre-symbol sum.

The residue $3$ is a primitive root modulo $17$. ChooseWriting $g=1+8k$ gives $k\equiv 13(\mathrm{mod}17)$, so one may takeIts Chinese-remainder components have orders $1$ and $16$, hence its order modulo $136$ is $16$.

A primitive root modulo $N$ is a unit of order $\phi (N)$; a nonzero quadratic residue modulo an odd prime $p$ is a square.

If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the subgroup of squares, contradicting ${\mathrm{ord}}_p(g)=p-1$.

Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd (k,10)=1$:Finally $2^{10}=1024\equiv 56\not\equiv 1(\mathrm{mod}121)$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.