Posting this self-answer, thanks to hints from Professor Dujella in the chat room.

First, we show that the assumption $G > 0$ leads to a contradiction.

It is clear that, under the assumption $G > 0$, $k \neq 1$ holds. (This is because $k = 1$ implies that $G = 0$.)

Then I obtain (from Equation $(4)$ in the original post) that
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{2(v_2 - v_1)}{u_1 - u_2} = \frac{q(tq^k - 2q + 2) - t}{(q - 1)(tq^k - 2q + 2)} = \frac{q}{q - 1} - \frac{t}{(q - 1)(tq^k - 2q + 2)}. \tag{1}$$

(Note that from Equation $(1)$ in this answer, we have $t \neq 0$.)

We obtain
$$\frac{1}{q^k (q - 1)} = \frac{t}{(q - 1)(tq^k - 2q + 2)}$$

which is equivalent to
$$0 = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{1}{q^k} - \frac{t}{tq^k - 2q + 2}\right) = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{tq^k - 2q + 2 - tq^k}{tq^k - 2q + 2}\right) = -\frac{2(q - 1)}{(q - 1)(tq^k - 2q + 2)}.$$
Now, since *a priori* we know that $q \geq 5$ holds, then we may cancel $q - 1$ from both numerator and denominator of the last fraction, to get
$$0 = -\frac{2}{tq^k - 2q + 2},$$
resulting in the contradiction
$$0 = -2.$$

(Note that $tq^k - 2q + 2 \neq 0$, because otherwise we would get
$$t = \frac{2(q - 1)}{q^k},$$
which implies that $q^k \mid 2(q - 1)$ (since $t$ must be an integer), contradicting $\gcd(2,q)=\gcd(q-1,q)=1$.)

Hence, we conclude that $G = 0$ must hold.

Now, we claim that:

**CLAIM:** $G = 0$ if and only if $k = 1$.

*Proof:* The implication $k = 1 \implies G = 0$ is obvious.

For the converse, assume that $G = 0$.

Then we obtain the system of equations
$$0 = u_1 - u_2 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot(tq^k - 2q + 2) \tag{2}$$
$$0 = v_2 - v_1 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$

Now, note that (referencing Equations $(2)$ and $(3)$ in this answer), we have
$$(2) \iff \left((q^k - q = 0) \lor (tq^k - 2q + 2 = 0)\right) \land (q \neq 1) \iff k = 1,$$
since $tq^k - 2q + 2 = 0$ contradicts the requirement that $t$ must be an integer.
$$(3) \iff \left((q^k - q = 0) \lor (q(tq^k - 2q + 2) - t = 0)\right) \land (q \neq 1) \iff \left((k = 1) \lor (t(q^{k+1} - 1) = 2q(q - 1))\right) \land (q \neq 1) \iff \Bigg((k = 1) \lor \left(1 < t = \frac{2q}{\sigma(q^k)} = \frac{2q}{q + 1} < 2\right)\Bigg) \land (q \neq 1) \iff k = 1,$$
which completes the proof.

Suppose $b = 2^t s$, for odd $s$. Let's look at the possible values of $2^a - 1 \pmod b$. According to the CRT, we can look at $2^a - 1 \pmod {2^t}, 2^a - 1 \pmod s$. It's easy to see that it must hit $(-1, 1)$ infinitely many times - $2^a \pmod s$ is periodic, and must hit $2$ infinitely many times, and for all $a \geq t$, $2^a \equiv 0 \pmod {2^t}$. Suppose from CRT we have $(-1, 1) \equiv v \pmod b$. It can be seen that $v^2\equiv 1\pmod b$, so if we use $va\mod b$ copies of $v$, they will sum to $a\pmod b$.

For example, consider $b=24, a=6$. So we have $2^t = 8$, $s = 3$. We have $2^a \equiv 2 \pmod 3 \iff a \equiv 1 \pmod 2$, so $2^3-1, 2^5-1,2^7-1, ...$ all have a value of $(-1, 1) \equiv 7 \pmod {24}$. We can use $7a \equiv 18 \pmod {24}$ such values, so we get
$$2^3-1 + 2^5-1 + 2^7-1 + \cdots + 2^{37}-1 \equiv 6 \pmod {24}$$
or
$n = 2^3 + 2^5 + \cdots + 2^{37} = 183251937960$.

## Best Answer

We have the following basic inequality : for all $m,n \geq 0$ , we have $$s_p(m) + s_p(n) \geq s_p(m+n)$$ (with equality if and only if there are no "carries"). This is not particularly difficult to prove in isolation.

Using this, even if $p$ doesn't divide $n$, we can see that the inequality holds. In this case, $v_p(n) = 0$, so all we need to prove is that $s_p(n-1)+s_p(n)+1 \geq s_p(2n)$, but this is clear once you notice that $$ s_p(2n) \leq s_p(n+1)+s_p(n-1) \leq s_p\left(n\right) +s_p(1)+ s_p(n-1) $$ and $s_p(1) = 1$.

We induct on $v_p(n)$ now. Suppose the fact is true for $v_p(n) = k$ with $k \geq 0$. We will assume that $v_p(n) = k+1$ so that $p|n$, and note that \begin{align} &s_p(n-1)+ s_p(n)+1-s_p(2n) \\ &= \left[s_p\left(\frac{n}{p}-1\right) + (p-1)\right]+s_p\left(\frac{n}{p}\right)+1 -s_p\left(\frac{2n}{p}\right)\\ &= (p-1) + \left[s_p\left(\frac{n}{p}-1\right) + s_p\left(\frac np\right) + 1 - s_p\left(\frac{2n}{p}\right)\right]\\ & \geq (p-1) + k(p-1) = (k+1)(p-1) \end{align}

as desired.