Let $a$ and $c$ be positive integers, and let $b$ be a digit. Determine all triples of numbers $(a, b, c)$ that satisfy the following conditions:

(1) $(a, b b b \ldots)^{2}=c, 777 \ldots$ (infinite decimal fractions);

(2) $\frac{c+a}{c-a}$ is an integer!
Let's rewrite the equation (1) by converting the decimal fractions to rational fractions. Then we get

$$
\left(a+\frac{b}{9}\right)^{2}=c+\frac{7}{9}
$$

By squaring and multiplying by 9, we obtain the following equation:

$$
9 a^{2}+2 a b+\frac{b^{2}}{9}=9 c+7
$$

Since $a$ and $c$ are positive integers, $\frac{b^{2}}{9}$ must also be an integer, so $b$ must be divisible by 3. However, $b$ cannot be 0 or 9, because then the left side of equation (1) would be an integer, while the right side would be a fraction. Therefore, $b$ can only be 3 or 6. Let's examine the second condition in these two cases.

I. If $b=3$, then substituting into equation (1b) and solving for $c$:

$$
c=\frac{3 a^{2}+2 a-2}{3}
$$

and

$$
\frac{c+a}{c-a}=\frac{3 a^{2}+5 a-2}{3 a^{2}-a-2}=\frac{(a+2)(3 a-1)}{(a-1)(3 a+2)}
$$

The difference between $3 a-1$ and $3 a+2$ is 3, so their common divisor must also divide 3. Moreover, $3 a-1$ is not divisible by 3, so these are relatively prime. Therefore, condition (2) can only be satisfied if $3 a+2$ is a divisor of the smaller $(a+2)$, which is not possible. In summary: there is no triplet that satisfies the conditions of the problem when $b=3$.

II. If $b=6$, then substituting into equation (1b) and solving for $c$:

$$
c=\frac{3 a^{2}+4 a-1}{3}
$$

and

$$
\frac{c+a}{c-a}=\frac{3 a^{2}+7 a-1}{3 a^{2}+a-1}=1+\frac{6 a}{3 a^{2}+a-1}
$$

For this to be an integer, it is necessary that

$$
6 a \geqq 3 a^{2}+a-1
$$

which gives us the quadratic inequality:

$$
3 a^{2}-5 a-1 \leqq 0
$$

The solution to this inequality is:

$$
\frac{5-\sqrt{37}}{6} \leqq a \leqq \frac{5+\sqrt{37}}{6}
$$

Only one positive integer falls into this interval, which is 1. If $a=1, b=6$, then $c=2$, and this triplet is the only one that satisfies all the conditions of the problem.