Given a prime $P$ with
$$P= r \cdot q+1$$
with $q$ prime as well.
I'm looking for a generator $g$ of the Schnorr subgroup with order $q$ which is small by value and has a inverse (to $\bmod P$) which is small as well:
$$|\{g^i, \forall i \in [0,P-1]\}| = q$$
$$g^q = 1 \bmod P$$
$$g < m \text{ } \land \text{ } g^{-1}<m$$
To find a Schnorr group generator $g$ in general we can pick a random value $h$ until $$ h^r \not= 1 \bmod P$$ $$\rightarrow \text{ } g = h^r \bmod P$$ To find $g^{-1}$ we can use the Extended Euclidean algorithm or just compute: $$ g^{-1} = g^{q-1} \bmod P$$
Questions:
Q1.) What is the best way to find $g,g^{-1}$ which are each less than value $m$? (edit: found some with step-by-step computation. Main interest now is Q3 (edit-end))
Q2.) How small can $m$ be?
Q3.) Would a small $g,g^{-1}$ have any impact to security?
Trial for Q1: Ways to find small $g, g^{-1}$
a.) pick random $v$ number and project to Schnorr group $v_s = v^r \bmod P$
(all values but $1$ of a Schnorr group are generators)
b.) compute inverse with EEA or with $v_s^{q-1} \bmod P$ (any way faster?)
c.) check if $g, g^{-1} < m$ . If not repeat a.)b.)c.)
Another way than random picks all the time would be an initial random pick for $g$, and a 2nd random value $w$ in this group to start from. Compute the inverse of both as well and traverse through the group with new generator $g^*_j$:
$$g_j^* = w \cdot (g)^j \bmod P$$
$${g_j^*}^{-1} = w^{-1} \cdot (g^{-1})^j \bmod P$$
$$w \cdot (g)^j \cdot w^{-1} \cdot (g^{-1})^j = 1 \bmod P$$
and test if $g^*_j$ and its inverse are smaller than $m$.
This computes the values faster but will it be faster in finding suitable generator? Can it be for certain values the results lay in a valley of bad values?
Trial for Q2: How small can $m$ be?
The product of $g, g^{-1}$ need to be at least $p+1$
That means $m \ge \sqrt{p-1}$
Is there another higher border for smallest $m$?
In general the equation need to hold: $$g \cdot g^{-1} = a\cdot P +1 \equiv 1 \bmod P $$ $$a \in [1,P-2]$$
Can the factorization of $(aP+1)$ be used to solve Q1 ($g,g^{-1}$ products of factor(combinations))?
Any way to guarantee its inside the Schnorr group?
Or just check again with $g^q = 1 \bmod P$?