Given the equation
$$
a^b\cdot b^c=c^a
$$
in positive integers $a$, $b$, and $c$.
[i](i)[/i] Prove that any prime divisor of $a$ divides $b$ as well.
[i](ii)[/i] Solve the equation under the assumption $b\ge a$.
[i](iii)[/i] Prove that the equation has infinitely many solutions.

[i](I. Voronovich)[/i]
**(i)** Prove that any prime divisor of \(a\) divides \(b\) as well.

1. Suppose \(p \mid a\) where \(p\) is a prime. This implies that \(a = p^m \cdot x\) for some integer \(m \geq 1\) and \(x\) not divisible by \(p\).
2. Since \(c^a\) is a multiple of \(p\), \(c\) must also be a multiple of \(p\). Thus, \(c = p^n \cdot y\) for some integer \(n \geq 1\) and \(y\) not divisible by \(p\).
3. Substitute \(a\) and \(c\) into the given equation:
   \[
   (p^m \cdot x)^b \cdot b^{(p^n \cdot y)} = (p^n \cdot y)^{(p^m \cdot x)}
   \]
4. Simplify the equation:
   \[
   p^{mb} \cdot x^b \cdot b^{p^n y} = p^{np^m x} \cdot y^{p^m x}
   \]
5. Equate the powers of \(p\) on both sides:
   \[
   mb + 0 = np^m x + 0
   \]
6. This implies:
   \[
   mb = np^m x
   \]
7. Since \(p \nmid x\) and \(p \nmid y\), \(p\) must divide \(b\). Hence, \(p \mid b\).

Thus, any prime divisor of \(a\) divides \(b\) as well. \(\blacksquare\)

**(ii)** Solve the equation under the assumption \(b \geq a\).

1. Given \(b \geq a\), we start with the equation:
   \[
   a^b \cdot b^c = c^a
   \]
2. Since \(b \geq a\), we have:
   \[
   a^{a+c} = a^a \cdot a^c \leq a^b \cdot b^c = c^a
   \]
3. This implies:
   \[
   a^{a+c} \leq c^a
   \]
4. Consider the cases \(a = c\), \(a > c\), and \(a < c\):

   - If \(a = c\), then:
     \[
     a^b \cdot b^a = a^a
     \]
     This simplifies to:
     \[
     a^b \cdot b^a = a^a
     \]
     Since \(a \neq 0\), we must have \(a = b = c = 1\).

   - If \(a > c\), then:
     \[
     a^{2c} < a^{a+c} \leq c^a < a^a
     \]
     This is a contradiction unless \(a = 1\).

   - If \(c > a\), then:
     \[
     (a^2)^a \leq a^{a+c} \leq c^a
     \]
     This implies:
     \[
     a^2 \leq c
     \]
     If \(a^n \leq c\) for some integer \(n\), then:
     \[
     a^{1+a^{n-1}} \leq c
     \]
     This leads to a contradiction unless \(a = 1\).

5. In all cases, we conclude that \(a = b = c = 1\).

Thus, the only solution when \(b \geq a\) is \(a = b = c = 1\). \(\boxed{(1, 1, 1)}\)

**(iii)** Prove that the equation has infinitely many solutions.

1. Consider the family of solutions \((a, b, c) = (t^t, t^{t-1}, t^t)\) for any positive integer \(t\).
2. Substitute these values into the equation:
   \[
   (t^t)^{t^{t-1}} \cdot (t^{t-1})^{t^t} = (t^t)^{t^t}
   \]
3. Simplify the left-hand side:
   \[
   t^{t \cdot t^{t-1}} \cdot t^{(t-1) \cdot t^t} = t^{t^t}
   \]
4. This simplifies to:
   \[
   t^{t^t} \cdot t^{t^t - t} = t^{t^t}
   \]
5. Since the exponents on both sides are equal, the equation holds true.

Thus, the equation has infinitely many solutions. \(\blacksquare\)