Let $a$, $b$, and $c$ be the number of direct roads between $A$ and $B$, $B$ and $C$, and $A$ and $C$, respectively. Let $x$, $y$, and $z$ be the number of roads from $A$ to $B$, $B$ to $C$, and $A$ to $C$, respectively (so $a = x$, $b = y$, $c = z$).

The total number of road links between $A$ and $B$ is the sum of direct roads and indirect roads via $C$:

- Direct: $a$
- Indirect: for each road from $A$ to $C$ and each road from $C$ to $B$, we get an indirect link. So $z \times y$.

Thus, $a + z y = 43$.

Similarly, the total number of road links between $B$ and $C$ is $b + x z = 29$.

The total number of road links between $A$ and $C$ is $c + x y$.

But since $a$, $b$, $c$ are the number of direct roads, and $x$, $y$, $z$ are the same, we can let $x = a$, $y = b$, $z = c$.

So:

- Road links between $A$ and $B$: $a + c b = 43$
- Road links between $B$ and $C$: $b + a c = 29$
- Road links between $A$ and $C$: $c + a b = ?$

Let $p = a$, $q = b$, $r = c$.

So:

$p + r q = 43$
$q + p r = 29$
$r + p q = ?$

We are to find the possible values of $r + p q$.

Let us solve for $p$, $q$, $r$.

From the first equation:
$p + r q = 43$
From the second:
$q + p r = 29$

Let us try to find integer solutions with $p, q, r \geq 1$.

Let us try small values for $r$.

Let $r = 1$:

First equation: $p + 1 \cdot q = 43 \implies p + q = 43$
Second: $q + p \cdot 1 = 29 \implies q + p = 29$

Contradicts.

Try $r = 2$:
First: $p + 2q = 43$
Second: $q + 2p = 29$

From first: $p = 43 - 2q$
Plug into second: $q + 2(43 - 2q) = 29 \implies q + 86 - 4q = 29 \implies -3q = -57 \implies q = 19$
Then $p = 43 - 2 \times 19 = 43 - 38 = 5$

So $p = 5$, $q = 19$, $r = 2$

Now, $r + p q = 2 + 5 \times 19 = 2 + 95 = 97$

Try $r = 3$:
First: $p + 3q = 43 \implies p = 43 - 3q$
Second: $q + 3p = 29 \implies q + 3(43 - 3q) = 29 \implies q + 129 - 9q = 29 \implies -8q = -100 \implies q = 12.5$
Not integer.

Try $r = 4$:
First: $p + 4q = 43 \implies p = 43 - 4q$
Second: $q + 4p = 29 \implies q + 4(43 - 4q) = 29 \implies q + 172 - 16q = 29 \implies -15q = -143 \implies q = 143/15 \approx 9.53$
Not integer.

Try $r = 5$:
First: $p + 5q = 43 \implies p = 43 - 5q$
Second: $q + 5p = 29 \implies q + 5(43 - 5q) = 29 \implies q + 215 - 25q = 29 \implies -24q = -186 \implies q = 186/24 = 7.75$
Not integer.

Try $r = 6$:
First: $p + 6q = 43 \implies p = 43 - 6q$
Second: $q + 6p = 29 \implies q + 6(43 - 6q) = 29 \implies q + 258 - 36q = 29 \implies -35q = -229 \implies q = 229/35 \approx 6.54$
Not integer.

Try $r = 7$:
First: $p + 7q = 43 \implies p = 43 - 7q$
Second: $q + 7p = 29 \implies q + 7(43 - 7q) = 29 \implies q + 301 - 49q = 29 \implies -48q = -272 \implies q = 272/48 = 5.67$
Not integer.

Try $r = 19$:
First: $p + 19q = 43 \implies p = 43 - 19q$
Second: $q + 19p = 29 \implies q + 19(43 - 19q) = 29 \implies q + 817 - 361q = 29 \implies -360q = -788 \implies q = 788/360 \approx 2.19$
Not integer.

Try $r = 43$:
First: $p + 43q = 43 \implies p = 43 - 43q$
Second: $q + 43p = 29 \implies q + 43(43 - 43q) = 29 \implies q + 1849 - 1849q = 29 \implies -1848q = -1820 \implies q = 1820/1848 \approx 0.985$
Not integer.

Try $r = 1$ again, but swap $p$ and $q$.

Alternatively, try $p = 1$:
First: $1 + r q = 43 \implies r q = 42$
Second: $q + r = 29$
So $r = 29 - q$
Plug into $r q = 42$:
$(29 - q) q = 42 \implies 29q - q^2 = 42 \implies q^2 - 29q + 42 = 0$
Solve quadratic:
$q = \frac{29 \pm \sqrt{29^2 - 4 \cdot 42}}{2} = \frac{29 \pm \sqrt{841 - 168}}{2} = \frac{29 \pm \sqrt{673}}{2}$
Not integer.

Try $q = 1$:
First: $p + r = 43$
Second: $1 + p r = 29 \implies p r = 28$
So $p + r = 43$, $p r = 28$
Try integer solutions:
$p = 1$, $r = 42$; $p r = 42$ (not 28)
$p = 2$, $r = 41$; $p r = 82$
$p = 4$, $r = 39$; $p r = 156$
$p = 7$, $r = 36$; $p r = 252$
$p = 14$, $r = 29$; $p r = 406$
$p = 28$, $r = 15$; $p r = 420$
No integer solution.

Try $p = 2$:
First: $2 + r q = 43 \implies r q = 41$
Second: $q + 2 r = 29 \implies q = 29 - 2 r$
Plug into $r q = 41$:
$r (29 - 2 r) = 41 \implies 29 r - 2 r^2 = 41 \implies 2 r^2 - 29 r + 41 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 328}}{4} = \frac{29 \pm \sqrt{513}}{4}$
Not integer.

Try $p = 3$:
First: $3 + r q = 43 \implies r q = 40$
Second: $q + 3 r = 29 \implies q = 29 - 3 r$
Plug into $r q = 40$:
$r (29 - 3 r) = 40 \implies 29 r - 3 r^2 = 40 \implies 3 r^2 - 29 r + 40 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 480}}{6} = \frac{29 \pm \sqrt{361}}{6} = \frac{29 \pm 19}{6}$
So $r = \frac{48}{6} = 8$, $r = \frac{10}{6} = 5/3$
So $r = 8$
Then $q = 29 - 3 \times 8 = 29 - 24 = 5$
Then $p = 3$

Now, $r + p q = 8 + 3 \times 5 = 8 + 15 = 23$

So possible values for $r + p q$ are $97$ and $23$.

Check for other integer solutions.

Try $p = 5$:
First: $5 + r q = 43 \implies r q = 38$
Second: $q + 5 r = 29 \implies q = 29 - 5 r$
Plug into $r q = 38$:
$r (29 - 5 r) = 38 \implies 29 r - 5 r^2 = 38 \implies 5 r^2 - 29 r + 38 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 760}}{10} = \frac{29 \pm \sqrt{81}}{10} = \frac{29 \pm 9}{10}$
So $r = \frac{38}{10} = 3.8$, $r = \frac{20}{10} = 2$
So $r = 2$
Then $q = 29 - 5 \times 2 = 29 - 10 = 19$
Then $p = 5$

Already found above.

Try $p = 8$:
First: $8 + r q = 43 \implies r q = 35$
Second: $q + 8 r = 29 \implies q = 29 - 8 r$
Plug into $r q = 35$:
$r (29 - 8 r) = 35 \implies 29 r - 8 r^2 = 35 \implies 8 r^2 - 29 r + 35 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 1120}}{16}$
$841 - 1120 = -279$, so no real solution.

Try $p = 10$:
First: $10 + r q = 43 \implies r q = 33$
Second: $q + 10 r = 29 \implies q = 29 - 10 r$
Plug into $r q = 33$:
$r (29 - 10 r) = 33 \implies 29 r - 10 r^2 = 33 \implies 10 r^2 - 29 r + 33 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 1320}}{20}$
$841 - 1320 = -479$, no real solution.

Try $p = 19$:
First: $19 + r q = 43 \implies r q = 24$
Second: $q + 19 r = 29 \implies q = 29 - 19 r$
Plug into $r q = 24$:
$r (29 - 19 r) = 24 \implies 29 r - 19 r^2 = 24 \implies 19 r^2 - 29 r + 24 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 1824}}{38}$
$841 - 1824 = -983$, no real solution.

Try $p = 23$:
First: $23 + r q = 43 \implies r q = 20$
Second: $q + 23 r = 29 \implies q = 29 - 23 r$
Plug into $r q = 20$:
$r (29 - 23 r) = 20 \implies 29 r - 23 r^2 = 20 \implies 23 r^2 - 29 r + 20 = 0$
Quadratic: $r = \frac{29 \pm \sqrt{841 - 1840}}{46}$
$841 - 1840 = -999$, no real solution.

Try $p = 43$:
First: $43 + r q = 43 \implies r q = 0$
So $r = 0$ or $q = 0$, but must be at least $1$.

Thus, the only integer solutions are $(p, q, r) = (5, 19, 2)$ and $(3, 5, 8)$, giving $r + p q = 97$ and $23$.

Therefore, the possible number of road links between $A$ and $C$ is $23$ or $97$.